Física para Ciencias e Ingeniería. Volumen II - Douglas C. Giancoli - 4ta edición [Solucionario]

Chapter 27 Magnetism

  IB0r 2 0 0  ˆj  sin0ˆj  cos2
sin ˆj  cos ˆi 0  IB0r sin 2 0  ˆi  cos0ˆi

 2IB0r sin0ˆj

The trigonometric identities sin 2    sin and cos2    cos are used to simplify the

solution.

10. We apply Eq. 27-3 to each circumstance, and solve for the magnetic field. Let  Bx ˆi  Byˆj  Bzkˆ .
 B
l
For the first circumstance,  l ˆi.

  ˆi ˆj kˆ
0
0  16.4Am Bzˆj  16.4Am Bykˆ  2.5ˆj N 
 FB  Il  B  8.2 A 2.0 m

Bx By Bz

By  0 ; 16.4Am Bz  2.5 N  Bz  2.5 N  0.1524 T ; Bx unknown
16.4Am

For the second circumstance, l  l ˆj.

  ˆi ˆj kˆ
2.0 m
FB  Il  B  8.2 A 0  0  2.5N ˆi  16.4Am Bxkˆ  2.5ˆi  5.0kˆ N 
0
Bx 0.1524 T

16.4Am Bx  5.0 N  Bx  5.0 N  0.3049 T
16.4Am
 

Thus B 
0.30 ˆi  0.15kˆ T .

11. We find the force along the wire by integrating the infinitesimal force from each path element (given

by Eq. 27-4) along an arbitrary path between the points a and b.
 b  b

     F   Id L B  I 
ˆidx  ˆjdy b ˆjdx  ˆidy  IB0 xˆj  yˆi

aa  B0kˆ  IB0
a

The resultant magnetic force on the wire depends on the displacement between the points a and b,

and not on the path taken by the wire. Therefore, the resultant force must be the same for the curved

path, as for the straight line path between the points.

12. The net force on the current loop is the sum of the z 
B
infinitesimal forces obtained from each current   
element. From the figure, we see that at each B
r d l  I
current segment, the magnetic field is   dl dF
perpendicular to the current. This results in a dF

force with only radial and vertical components.

By symmetry, we find that the radial force

components from segments on opposite sides of the loop cancel. The net force then is purely

vertical. Symmetry also shows us that each current element contributes the same magnitude of force.

  IBrkˆ  d l  I B sin  kˆ 2 r  2 IB r2 kˆ
r2  d2
F   Id L  B

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

13. The maximum magnetic force as given in Eq. 27-5b can be used since the velocity is perpendicular
to the magnetic field.

   Fmax  qvB  1.60 1019C 8.75 105 m s 0.45T  6.31014 N

By the right hand rule, the force must be directed to the North .

14. The magnetic force will cause centripetal motion, and the electron will move in a clockwise circular

path if viewed in the direction of the magnetic field. The radius of the motion can be determined.

    Fmax
 qvB  m v2  r  mv  9.111031kg 1.70 106 m s  2.02 105 m
r qB
1.60 1019 C 0.480T

15. In this scenario, the magnetic force is causing centripetal motion, and so must have the form of a

centripetal force. The magnetic force is perpendicular to the velocity at all times for circular motion.

    Fmax
 qvB  m v2  B  mv  6.6 1027 kg 1.6 107 m s  1.8T
r qr
2 1.60 1019 C 0.18 m

16. Since the charge is negative, the answer is the OPPOSITE of the result given from the right hand rule
applied to the velocity and magnetic field.
(a) left
(b) left
(c) upward
(d) inward into the paper
(e) no force
(f) downward

17. The right hand rule applied to the velocity and magnetic field would give the direction of the force.
Use this to determine the direction of the magnetic field given the velocity and the force.
(a) downward
(b) inward into the paper
(c) right

18. The force on the electron due to the electric force must be the same magnitude as the force on the
electron due to the magnetic force.

FE  FB  qE  qvB  v  E  8.8 103 V m  1.173106 m s  1.2 106 m s
B 7.5 103T

If the electric field is turned off, the magnetic force will cause circular motion.

    FB
 qvB  m v2  r  mv  9.111031kg 1.173 106 m s  8.9 104 m
r qB 1.60 1019 C 7.5 103T

19. (a) The velocity of the ion can be found using energy conservation. The electrical potential energy

of the ion becomes kinetic energy as it is accelerated. Then, since the ion is moving
perpendicular to the magnetic field, the magnetic force will be a maximum. That force will
cause the ion to move in a circular path.

E  Einitialfinal  qV  1 mv2  v 2qV
2 m

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202

Chapter 27 Magnetism

Fmax  qvB  m v2 
r

r  mv  m 2qV    2
6.6 1027 kg 2700 V
m 1 2mV 1  3.1102 m
  2 1.60 1019 C
qB qB B
q 0.340T

(b) The period can be found from the speed and the radius. Use the expressions for the radius and
the speed from above.

2 1 2mV
Bq

2qV
   
v  2 r T  2 r   2 m  2 6.61027 kg  3.8107 s
T v qB
2 1.601019 C 0.340 T

m

20. The velocity of each charged particle can be found using energy conservation. The electrical
potential energy of the particle becomes kinetic energy as it is accelerated. Then, since the particle is
moving perpendicularly to the magnetic field, the magnetic force will be a maximum. That force
will cause the ion to move in a circular path, and the radius can be determined in terms of the mass
and charge of the particle.

E  Einitial final  qV  1 mv2  v 2qV
2 m

r  mv  m 2qV

Fmax  qvB  m v2  m 1 2mV
r  q
qB qB B

1 2mdV md

rd  B qd  mp  2 2  rd  2rp
rp 1 2mpV qd 

1

B qp qp

1 2mV m

r  B q  mp  4 2  r  2rp
rp 1 2mpV q 

2

B qp qp

21. (a) From Example 27-7, we have that r  mv , and so v rqB . The kinetic energy is given by
qB m

K  1 mv2  1 m  rqB 2  r2q2B2 and so we see that K  r2 .
2 2 m 2m



(b) The angular momentum of a particle moving in a circular path is given by L  mvr . From

Example 27-7, we have that r  mv , and so v  rqB . Combining these relationships gives
qB m

L  mvr  m rqB r  qBr2 .

m

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22. The force on the electron is given by Eq. 27-5a.

 qv  ˆi ˆj kˆ
FB B
   e 7.0 104 m s 6.0 104 m s 0  e 4.2  4.8 104 Tm s kˆ

0.80 T 0.60 T 0

     1.60 1019 C 0.6 104 Tm s kˆ  9.6 1016 N kˆ  11015 N kˆ

23. The kinetic energy of the proton can be used to find its velocity. The magnetic force produces
centripetal acceleration, and from this the radius can be determined.

K  1 mv2  v 2K qvB  mv2  r  mv
2 m r qB

mv m 2K 2Km      2 6.0 106eV 1.60 1019 J eV 1.67 1027 kg
qB qB 
r   m   1.60 1019C 0.20T  1.8 m
qB

24. The magnetic field can be found from Eq. 27-5b, and the direction is found from the right hand rule.
Remember that the charge is negative.

B  Fmax  8.2 1013 N
qv 1.60 1019 C 2.8 106 m s
   Fmax  qvB   1.8T

The direction would have to be East for the right hand rule, applied to the velocity and the
magnetic field, to give the proper direction of force.

25. The total force on the proton is given by the Lorentz equation, Eq. 27-7.

   ˆi ˆj kˆ 
   E v B 3.0 103 m s
 
FB  q  0.38 T 
   e 3.0ˆi  4.2ˆj 103 V m  6.0 103 m s 5.0 103 m s

 0.45T 0 

      1.60 1019C  3.0ˆi  4.2ˆj  1.9ˆi  2.25ˆj  0.93kˆ  103 N C

    1.60 1019C  4.9ˆi  6.45ˆj  0.93kˆ  103 N C

  7.84 1016 ˆi  1.03 1015ˆj  1.49 1016 kˆ N C

   0.78ˆi 1.0ˆj  0.15kˆ  1015 N

26. The force on the electron is given by Eq. 27-5a. Set the force expression components equal and

solve for the velocity components.

 qv  ˆi ˆj kˆ
FB B
   Fxˆi  Fyˆj  e vx vy vz  evyBzˆi  e vxBz  ˆj 

0 0 Bz

vy   Fx 3.8 1013 N  2.8 106 m s
eBz
1.60 1019 C 0.85T
 Fx  evyBz  

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204

Chapter 27 Magnetism

vx Fy 2.7 1013 N 2.0 106
eBz
1.60 1019 C 0.85T
 Fy  evxBz     m s

 v   2.0ˆi  2.8ˆj 106 m s

Notice that we have not been able to determine the z component of the electron’s velocity.

27. The kinetic energy of the particle can be used to find its velocity. The magnetic force produces
centripetal acceleration, and from this the radius can be determined. Inserting the radius and velocity
into the equation for angular momentum gives the angular momentum in terms of the kinetic energy
and magnetic field.

K  1 mv2  v 2K qvB  mv2  r  mv
2 m r qB

 m 2K 
 
L  mvr  m 2K  m   2mK

m  qB  qB

From the equation for the angular momentum, we see that doubling the magnetic field while keeping

the kinetic energy constant will cut the angular momentum in half.

Lfinal  1 Linitial
2

28. The centripetal force is caused by the magnetic field, and is given by Eq. 27-5b.

F  qvB sin  qv B  m v2 
 

r

  r
 mv  9.111031kg 3.0 106 m s sin 45o  4.314 105 m  4.3 105 m

1.60 1019C 0.28T
 qB

The component of the velocity that is parallel to the magnetic field is unchanged, and so the pitch is

that velocity component times the period of the circular motion.

T 2 r 2 mv  2 m
  

qB

v v qB


     p
 vT  v cos 45o  2 m   3.0 106 m s cos 45o 2 9.111031kg 2.7 104 m
 qB 
1.60 1019 C 0.28T 

29. (a) For the particle to move upward the magnetic force
must point upward, by the right hand rule we see that
the force on a positively charged particle would be
downward. Therefore, the charge on the particle must

be negative.

(b) In the figure we have created a right triangle to relate

the horizontal distance l, the displacement d, and the
radius of curvature, r. Using the Pythagorean theorem
we can write an expression for the radius in terms of
the other two distances.

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

r2  r  d2  l2  r  d2  l2
2d

Since the momentum is perpendicular to the magnetic field, we can solve for the momentum by

relating the maximum force (Eq. 27-5b) to the centripetal force on the particle.

 Fmax
 qvB0  mv2  p  mv  qB0r  qB0 d 2  l2
r 2d

30. In order for the path to be bent by 90 within a distance d, the radius of curvature must be less than
or equal to d. The kinetic energy of the protons can be used to find their velocity. The magnetic
force produces centripetal acceleration, and from this, the magnetic field can be determined.

K  1 mv2  v 2K qvB  mv2  B  mv
2 m r qr

mv m 2K  2Km 1/ 2
ed  e2d 2 
B   m 
ed

31. The magnetic force will produce centripetal acceleration. Use that relationship to calculate the

speed. The radius of the Earth is 6.38 106 km , and the altitude is added to that.

    FB
 qvB  m v2  v  qrB  1.60 1019 C 6.385 106 m 0.50 104 T  1.3 108 m s
r m 238 1.66 1027 kg

Compare the size of the magnetic force to the force of gravity on the ion.

   FB  qvB 
  Fg mg
1.60 1019 C 1.3 108 m s 0.50 104 T  2.3108
238 1.66 1027 kg 9.80 m s2

Yes, we may ignore gravity. The magnetic force is more than 200 million times larger than gravity.

32. The magnetic force produces an acceleration that is perpendicular to the original motion. If that
perpendicular acceleration is small, it will produce a small deflection, and the original velocity can
be assumed to always be perpendicular to the magnetic field. This leads to a constant perpendicular
acceleration. The time that this (approximately) constant acceleration acts can be found from the

original velocity v and the distance traveled l. The starting speed in the perpendicular direction will
be zero.

F  ma  qvB  a  qvB
m

18.5 109 C 5.00 105T 1.00 103 m 2

2 3.40 103kg 155m s
   d
 
 v0t  1 at 2  1 qvB  l 2  qBl2 
2  2 m  v  2mv

 8.8 107 m
This small distance justifies the assumption of constant acceleration.

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206

Chapter 27 Magnetism

33. (a) In the magnetic field, the proton will move along an arc of a 8
circle. The distance x in the diagram is a chord of that circle,

and so the center of the circular path lies on the perpendicular

bisector of the chord. That perpendicular bisector bisects the 6 7
central angle of the circle which subtends the chord. Also recall
that a radius is perpendicular to a tangent. In the diagram, 5
4
1  2 because they are vertical angles. Then 2  4 , because r 3 2

they are both complements of 3 , so 1  4 . We have 4  5
since the central angle is bisected by the perpendicular bisector

of the chord. 5  7 because they are both complements of 6 , 1

and 7  8 because they are vertical angles. Thus

1  2  4  5  7  8 , and so in the textbook diagram, the angle at which the proton leaves

is   45o .

(b) The radius of curvature is given by r  mv , and the distance x is twice the value of r cos .
qB

   x  2r cos  2 mv cos  2
 qB
1.67 1027 kg 1.3 105 m s cos 45o  2.3 103 m

1.60 1019 C 0.850T

34. (a) Since the velocity is perpendicular to the magnetic field,
the particle will follow a circular trajectory in the x-y plane
of radius r. The radius is found using the centripetal
acceleration.

qvB  mv2  r  mv
r qB

From the figure we see that the distance  is the chord
distance, which is twice the distance r cos. Since the

velocity is perpendicular to the radial vector, the initial direction and the angle  are
complementary angles. The angles  and  are also complementary angles, so   30.

  2r cos  2mv0 cos 30  3 mv0
qB0 qB0

(b) From the diagram, we see that the particle travels a circular path, that is 2 short of a complete
circle. Since the angles  and  are complementary angles, so   60. The trajectory distance

is equal to the circumference of the circular path times the fraction of the complete circle.

Dividing the distance by the particle speed gives t.

t  l  2 r  360  260    2 mv0  2   4 m
v0 v0   v0 qB0  3  3qB0
360

(eTanh)eergwWyoirskgrieUvqeunirbeydμEbyqB.a2n7e-1x2te,rUnμalBagμenBti.s equal to the change in potential energy. The potential
final initial μB  μ B 
 NIAB cosinitial  cosfinal
initial final
       35.

 NIAB cos 0  cos180  2NIAB

(b) W  NIAB cosinitial  cosfinal  NIAB cos 90  cos 90  0

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

36. With the plane of the loop parallel to the magnetic field, the torque will be a maximum. We use Eq.

27-9.

  NIAB sin  B    0.185 m N  3.32 T
NIAB sin
1 4.20 A 0.0650 m2 sin 90

37. (a) The torque is given by Eq. 27-9. The angle is the angle between the B-field and the

perpendicular to the coil face.

  0.180 m 2 
 2  
   
 NIAB sin  12  7.10 A 5.50 105T sin 24  4.85 105 mN

(b) In Example 27-11 it is stated that if the coil is free to turn, it will rotate toward the orientation so

that the angle is 0. In this case, that means the north edge of the coil will rise, so that a

perpendicular to its face will be parallel with the Earth’s magnetic field.

38. The magnetic dipole moment is defined in Eq. 27-10 as   NIA . The number of turns, N , is 1.

The current is the charge per unit time passing a given point, which on the average is the charge on

the electron divided by the period of the circular motion, I  e T . If we assume the electron is

moving in a circular orbit of radius r, then the area is  r2 . The period of the motion is the

circumference of the orbit divided by the speed, T  2 r v . Finally, the angular momentum of an

object moving in a circle is given by L  mrv . Combine these relationships to find the magnetic
moment.

  NIA  e  r2  e  r2  e r2v  erv  emrv  e mrv  e L
T 2 r v 2 r 2 2m 2m 2m

39. (a) The magnetic moment of the coil is given by Eq. 27-10. Since the current flows in the

clockwise direction, the right hand rule shows that the magnetic moment is down, or in the

negative z-direction.

 μ   0.22 m 2 kˆ  4.334 kˆ Am2  4.3kˆ Am2
 NIA  157.6 A  2



(b) We use Eq. 27-11 to find the torque on the coil.
     μ
  B  4.334 kˆ Am2  0.55ˆi  0.60ˆj  0.65kˆ T 2.6ˆi  2.4ˆj mN

    (c)
We use Eq.μ2B7-12tof4i.n3d34thkeˆ potential energy of the coil.
U Am2 0.55ˆi  0.60ˆj  0.65kˆ
T 4.334Am2 0.65 T

 2.8 J

40. To find the total magnetic moment, we divide the rod into infinitesimal pieces of

thickness dy. As the rod rotates on its axis the charge in each piece, Q d  dy, + 

creates a current loop around the axis of rotation. The magnitude of the current dy +
is the charge times the frequency of rotation,  2 . By integrating the
+
infinitesimal magnetic moments from each piece, we find the total magnetic +
moment. +

 μ dμ  d   Q  Q d y2dy  Qd 2 y+
    AdI   y2 2 d dy   2d 6 .
0  0 +
+

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Chapter 27 Magnetism

41. From Section 27-5, we see that the torque is proportional to the current, so if the current drops by
12%, the output torque will also drop by 12%. Thus the final torque is 0.88 times the initial torque.

42. In Section 27-6, it is shown that the deflection of the galvanometer needle is proportional to the

product of the current and the magnetic field. Thus if the magnetic field is decreased to 0.860 times

its original value, the current must be increased by dividing the original value by 0.860 to obtain the

same deflection.

 
   IB initial IB I Binitial initial 63.0 A Binitial  78.8 A
final I  final 0.800Binitial

Bfinal

43. From the galvanometer discussion in Section 27-6, the amount of deflection is proportional to the

ratio of the current and the spring constant:   I . Thus if the spring constant decreases by 15%,
k

the current can decrease by 15% to produce the same deflection. The new current will be 85% of the

original current.

 Ifinal  0.85Iinitial  0.85 46A  39A

44. Use Eq. 27-13.

q  E  260 V m  1.5 105 C kg
m B2r 0.46T2 0.0080 m

45. The force from the electric field must be equal to the weight.

    qE
 ne  V   mg  n  mgd  3.31015 kg 9.80 m s2 0.010 m  5.94  6 electrons
 d  eV 1.60 1019 340 V

46. (a) Eq. 27-14 shows that the Hall emf is proportional to the magnetic field perpendicular to the
conductor’s surface. We can use this proportionality to determine the unknown resistance.

Since the new magnetic field is oriented 90 to the surface, the full magnetic field will be used
to create the Hall potential.

eH   B  B  eH  B  63 mV  0.10 T   0.53T
eH B eH 12 mV

(b) When the field is oriented at 60 to the surface, the magnetic field, B sin 60, is used to create

the Hall potential.

B sin 60  eH  B  B  63 mV  0.10 T   0.61T
eH 12 mV
sin 60

47. (a) We use Eq. 27-14 for the Hall Potential and Eq. 25-13 to write the current in terms of the drift

velocity.

KH  eH  vd Bd  1
IB ent
en td  vd  B

(b) We set the magnetic sensitivities equal and solve for the metal thickness.

1 ns 31022 m3
enmtm nm 11029 m3
 1

ensts
  tm  ts  0.15 103 m  51011 m

This is less than one sixth the size of a typical metal atom.

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(c) Use the magnetic sensitivity to calculate the Hall potential.

KH IB IB 100 mA0.1T  14 mV  10 mV
ent
1.6 1019 C 31022 m3 0.15103 m
   eH  

48. (a) We find the Hall field by dividing the Hall emf by the width of the metal.

EH  eH  6.5 V  2.167 104 V/m  2.2 104 V/m
d 0.03 m

(b) Since the forces from the electric and magnetic fields are balanced, we can use Eq. 27-14 to

calculate the drift velocity.

vd  EH  2.167 104 V/m  2.709 104 m s 2.7 104 m/s
B 0.80 T

(c) We now find the density using Eq. 25-13.

I 42 A
eAvd
6.80104 m 0.03m 2.709 104 m/s
    n 
1.6 1019C

 4.7 1028electrons/m3

49. We find the magnetic field using Eq. 27-14, with the drift velocity given by Eq. 25-13. To

determine the electron density we divide the density of sodium by its atomic weight. This gives the

number of moles of sodium per cubic meter. Multiplying the result by Avogadro’s number gives

the number of sodium atoms per cubic meter. Since there is one free electron per atom, this is also

the density of free electrons.

B  eH  eH  eH net  eH et  NA 
vd d I I  mA 
 I  d  
 ne(td ) 
 

      1.86106 V 1.601019C 1.30103m 0.971 1000 kg/m3 6.022 1023e/mole
 12.0 A
0.02299 kg/mole

 0.820T

50. (a) The sign of the ions will not change the magnitude of the Hall emf, but will

determine the polarity of the emf .

(b) The flow velocity corresponds to the drift velocity in Eq. 27-14.

 EH  vBd 
 v  EH  0.13 103 V  0.56 m s
Bd
 0.070 T   0.0033 m 

51. The magnetic force on the ions causes them to move in a circular path, so the magnetic force is a
centripetal force. This results in the ion mass being proportional to the path’s radius of curvature.

qvB  m v2 r  m  qBr v  m r  qB v  constant  76 u 22.8 cm

m21.0  76 u  m21.0  70 u m21.6  76 u  m21.6  72 u
21.0 cm 22.8 cm 21.6 cm 22.8 cm

m21.9  76 u  m21.9  73 u m22.2  76 u  m22.2  74 u
21.9 cm 22.8 cm 22.2 cm 22.8 cm

The other masses are 70 u, 72 u, 73 u, and 74 u .

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210

Chapter 27 Magnetism

52. The velocity of the ions is found using energy conservation. The electrical potential energy

of the ions becomes kinetic energy as they are accelerated. Then, since the ions move

perpendicularly to the magnetic field, the magnetic force will be a maximum. That force will cause

the ions to move in a circular path.

qvB  mv2  v  qBR qV  1 mv2  1 m  qBR 2  q2B2R2  m  qR2 B2
Rm 2 2  m  2m 2V

53. The location of each line on the film is twice the radius of curvature of the ion. The radius of

curvature can be found from the expression given in Section 27-9.

m  qBBr  r  mE  2r  2mE
E qBB qBB

   2 12 1.67 1027 kg 2.48 104 V m  1.8467 102 m
 2r12 
1.60 1019 C 0.58 T 2

2r13  2.0006 102 m 2r14  2.1545 102 m

The distances between the lines are

2r13  2r12  2.0006 102 m  1.8467 102 m  1.539 103 m  1.5 103 m

2r14  2r13  2.1545 102 m  2.0006 102 m  1.539 103 m  1.5 103 m

If the ions are doubly charged, the value of q in the denominator of the expression would double, and
so the actual distances on the film would be halved. Thus the distances between the lines would also
be halved.

2r13  2r12  1.0003 102 m  9.2335 103 m  7.695 104 m  7.7 104 m

2r14  2r13  1.07725 102 m  1.0003 102 m  7.695 104 m  7.7 104 m

54. The particles in the mass spectrometer follow a semicircular path as shown in Fig. 27-33. A particle
has a displacement of 2r from the point of entering the semicircular region to where it strikes the
film. So if the separation of the two molecules on the film is 0.65 mm, the difference in radii of the
two molecules is 0.325 mm. The mass to radius ratio is the same for the two molecules.

qvB  m v2 r  m  qBr v  m r  constant

 m   m   28.0106 u 28.0134 u  r  3.251m  3.3 m
 r CO   r  N 2 r  r  3.25 104 m

55. Since the particle is undeflected in the crossed fields, its speed is given by Eq. 27-8. Without the

electric field, the particle will travel in a circle due to the magnetic force. Using the centripetal
acceleration, we can calculate the mass of the particle. Also, the charge must be an integer multiple
of the fundamental charge.

qvB  mv2 
r
   m 2
 qBr  qBr  neB2r  n 1.60 1019 C 0.034 T 0.027 m  n 3.31027 kg  n 2.0 u
v E
E / B 1.5 103 V/m

The particle has an atomic mass of a multiple of 2.0 u. The simplest two cases are that it could be a

hydrogen-2 nucleus (called a deuteron), or a helium-4 nucleus (called an alpha particle): 2 H, 4 He .
1 2

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56. The radius and magnetic field values can be used to find the speed of the protons. The electric field
is then found from the fact that the magnetic force must be the same magnitude as the electric force
for the protons to have straight paths.

qvB  m v2 r  v  qBr m FE  FB  qE  qvB 

   E  vB  qB2r m 
1.60 1019 C 0.625T2 5.10 102 m  1.91106 V m

1.67 1027 kg

The direction of the electric field must be perpendicular to both the velocity and the magnetic field,

and must be in the opposite direction to the magnetic force on the protons.

57. The magnetic force produces centripetal acceleration.

p 3.8 1016 kgm s
qr 1.60 1019 C 1.0 103 m
  qvB  m v2 r B
 mv  p  qBr     2.4 T

The magnetic field must point upward to cause an inward-pointing (centripetal) force that steers the
protons clockwise.

58. The kinetic energy is used to determine the speed of the particles, and then the speed can be used to
determine the radius of the circular path, since the magnetic force is causing centripetal acceleration.

2K qvB  mv2 mv m 2K 2m K
m r qB qB
K  1 mv2  v  r   m 
2 qB

2mp K

rp  qB  mp  1.67 1027 kg  42.8
re 2me K me 9.11  1031kg

qB

59. (a) There will be one force on the rod, due to the magnetic force on the charge carriers in the rod.

That force is of magnitude FB  IdB , and by Newton’s second law is equal to the mass of the

rod times its acceleration. That force is constant, so the acceleration will be constant, and
constant acceleration kinematics can be used.

Fnet  FB  IdB  ma  a  IdB  v  v0  v  v IdB t
m tt m

(b) Now the net force is the vector sum of the magnetic force and the force of kinetic friction.
Fnet  FB  Ffr  IdB  kFN  IdB  kmg  ma 

a  IdB  k g  v  v0  v  v   IdB  k g  t
m t t  m 

(c) Using the right hand rule, we find that the force on the rod is to the east, and the rod moves east.

60. Assume that the magnetic field makes an angle  with respect to the  
vertical. The rod will begin to slide when the horizontal magnetic force B

 IBl cos  is equal to the maximum static friction sFN . Find the FB

normal force by setting the sum of the vertical forces equal to zero. See 
the free body diagram. Ffr FN

FB sin  FN  mg  0  FN  mg  FB sin  mg  I lB sin mg

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Chapter 27 Magnetism

IBl cos  sFN =s mg  I lB sin   B  I l  s smg cos 
sin 

We find the angle for the minimum magnetic field by setting the derivative of the magnetic field

with respect to the angle equal to zero and solving for the angle.

dB  0  smg s cos  sin     tan-1s  tan-10.5  26.6
d I l s sin  cos 2

 B
 I l s smg cos   0.50.40 kg 9.80 m/s2  0.22 T
sin  36 A0.22 m0.5sin 26.6  cos26.6

The minimum magnetic field that will cause the rod to move is 0.22 T at 27 from the vertical.

61. The magnetic force must be equal in magnitude to the weight of the electron.

  mg  qvB
 v  mg  9.111031kg 9.80 m s2  1.1106 m s
qB 1.60 1019C 0.50 104 T

The magnetic force must point upwards, and so by the right hand rule and the negative charge of the
electron, the electron must be moving west .

62. The airplane is a charge moving in a magnetic field. Since it is flying perpendicular to the magnetic
field, Eq. 27-5b applies.

   Fmax  qvB  1850 106C 120 m s 5.0 105T  1.1105 N

63. The maximum torque is found using Eq. 27-9 with sin  1 . Set the current equal to the voltage

divided by resistance and the area as the square of the side length.

  NIAB  N  V  l2 B  20  9.0 V  0.050 m 2  0.020 T   3.8104 mN
 R   24  

64. The speed of the electrons is found by assuming the energy supplied

by the accelerating voltage becomes kinetic energy of the electrons.

We assume that those electrons are initially directed horizontally, vx y
and that the television set is oriented so that the electron velocity is
x

perpendicular to the Earth’s magnetic field, resulting in the largest

possible force. Finally, we assume that the magnetic force on the

electrons is small enough that the electron velocity is essentially

perpendicular to the Earth’s field for the entire trajectory. This results in a constant acceleration for

the electrons.

(a) Acceleration:

U  Kinitial final  eV  1 mv 2  vx  2eV
2 x m

Deflection:

time in field: xfield  v tx field  tfield  xfield
vx

qvx BEarth e 2eV BEarth 2e3V
m m m3
Fy  qvx BEarth  ma y  ay    BEarth
m

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y  1 ayt2  1 2e3V BEarth  xfield 2  1  2e3VBEarth x 2 m
2 2 m3  vx  2 2eV
  m3

 e 1.60 1019 C
8 9.11 1031 kg 2.0 103 V
8mV
    
BEarth x 2  5.0 105T 0.18 m2

 5.37 103 m  5.4 mm

 e 1.60 1019 C

8mV 8 9.111031kg 28, 000 V
   (b)
y  BEarth x 2  5.0 105 T 0.18 m2

 1.4 103 m

Note that the deflection is significantly smaller than the horizontal distance traveled, and so the
assumptions made above are verified.

65. From Fig. 27-22 we see that when the angle  is positive, the torque is negative. The magnitude of

the torque is given by Eq. 27-9. For small angles we use the approximation sin  . Using Eq. 10-
14, we can write the torque in terms of the angular acceleration, showing that it is a harmonic
oscillator.

  NIAB sin  IabB  IM      IabB   2
 IM

We obtain the period of motion from the angular frequency, using T  2  . First we determine the

moment of inertia of the loop, as two wires rotating about their centers of mass and two wires
rotating about an axis parallel to their lengths.

IM  2 1  2a b 2b m  b2   2  2a a 2b m   b 2  3a  bmb2
12         2 12a  b
 


T  2  2 IM  2 mb2 3a  b   mb3a  b
 NIabB 12a  b NIabB 3a  b NIaB

66. (a) The frequency of the voltage must match the frequency of circular motion of the particles, so
that the electric field is synchronized with the circular motion. The radius of each circular orbit

is given in Example 27-7 as r  mv . For an object moving in circular motion, the period is
qB

given by T  2 r , and the frequency is the reciprocal of the period.
v

T  2 r  f  v  v  Bq
v 2 r 2 mv 2 m

qB

In particular we note that this frequency is independent of the radius, and so the same frequency

can be used throughout the acceleration.
(b) For a small gap, the electric field across the gap will be approximately constant and uniform as

the particles cross the gap. If the motion and the voltage are synchronized so that the maximum

voltage occurs when the particles are at the gap, the particles receive an energy increase of

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214

Chapter 27 Magnetism

K  qV0 as they pass each gap. The energy gain from one revolution will include the passing

of 2 gaps, so the total kinetic energy increase is 2qV0 .

(c) The maximum kinetic energy will occur at the outside of the cyclotron.

0.50 m 2 1.60 1019 C 2
1.67 1027 kg
     Kmax
 mv1 2  1 m  rmax qB 2  1 r2 q2 B2  1 0.60T 2
2  m 2 2
2 max  max

m

 6.898  1013 J  1eV J   1MeV   4.3 MeV
 1.60 1019   106 eV 

67. The protons will follow a circular path as they move through the

region of magnetic field, with a radius of curvature given in Example

27-7 as r  mv . Fast-moving protons will have a radius of curvature l
qB rr

that is too large and so they will exit above the second tube. 
Likewise, slow-moving protons will have a radius of curvature that is
too small and so they will exit below the second tube. Since the exit
velocity is perpendicular to the radius line from the center of
curvature, the bending angle can be calculated.

sin  l 
r

     sin1 l  sin1 lqB  sin1
  r mv
5.0 102 m 1.60 1019 C 0.38T  sin1 0.214  12

1.67 1027 kg 0.85 107 m s

68. (a) The force on each of the vertical wires in the loop is
perpendicular to the magnetic field and is given by Eq. 27-1,
with   90. When the face of the loop is parallel to the

magnetic field, the forces point radially away from the axis.
This provides a tension in the two horizontal sides. When the
face of the loop is perpendicular to the magnetic field, the force
on opposite vertical wires creates a shear force in the horizontal
wires. From Table 12-2, we see that the tensile and shear
strengths of aluminum are the same, so either can be used to
determine the minimum strength. We set tensile strength multiplied by the cross-sectional area
of the two wires equal the tensile strength multiplied by the safety factor and solve for the wire
diameter.

2 10I lB 1015.0 A0.200 m1.35 T

  F A  200 106 N/m2
F   d   10  I lB   d 2  2
 2
 A

 5.0777 104 m  0.508 mm

(b) The resistance is found from the resistivity using Eq. 25-3.

 R   l 4  0.200 m 
A  2.65 108 m  0.105
5.0777 10-4 2
  2 m 

 

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215

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

69. The accelerating force on the bar is due to the magnetic force on the current. If the current is
constant, the magnetic force will be constant, and so constant acceleration kinematics can be used.

v2  v02  2ax  a  v2  0  v2
2x 2x

m  v2 
 2x 
 Fnet  ma  IdB
 I  ma  dB  mv2  1.5 103 kg 25m s2  1.1A
dB 2xdB 21.0 m 0.24 m 1.8T

Using the right hand rule, for the force on the bar to be in the direction of the acceleration shown in

Fig. 27-53, the magnetic field must be down .

70. (a) For the beam of electrons to be undeflected, the magnitude of the magnetic force must equal the
magnitude of the electric force. We assume that the magnetic field will be perpendicular to the
velocity of the electrons so that the maximum magnetic force is obtained.

FB  FE  qvB  qE  B  E  8400 V m  1.75 103T  1.8 103T
v 4.8 106 m s

(b) Since the electric field is pointing up, the electric force is down. Thus the magnetic force must

be up. Using the right hand rule with the negative electrons, the magnetic field must be out of

the plane of the plane formed by the electron velocity and the electric field.
(c) If the electric field is turned off, then the magnetic field will cause a centripetal force, moving

the electrons in a circular path. The frequency is the reciprocal of the period of the motion.

qvB  mv2  v  qBr
rm

qBr
  f
1  v  m  qB  qE  2 1.60 1019 C 8400 V m  4.9 107 Hz
T 2 r 2 r 2 m 2 mv
9.111031kg 4.8 106 m s

71. We find the speed of the electron using conservation
of energy. The accelerating potential energy becomes
the kinetic energy of the electron.

eV  1 mv 2  v 2eV
2 m

Upon entering the magnetic field the electron is
traveling horizontally. The magnetic field will cause

the path of the electron to rotate an angle  from the
horizontal. While in the field, the electron will travel
a horizontal distance d and a vertical distance h0.
Using the Pythagorean theorem, and trigonometric
relations, we can write three equations which relate

the unknown parameters, r, h0, and .

tan = h  h0 sin  d r2  d 2  r  h0 2  h0  r  r2  d 2
l  d r

These three equations can be directly solved, for the radius of curvature. However, doing so requires
solving a 3rd order polynomial. Instead, we can guess at a value for h0, such as 1.0 cm. Then we use

the tangent equation to calculate an approximate value for . Then insert the approximate value into

the sine equation to solve for r. Finally, inserting the value of r into the third equation we solve for

h0. We then use the new value of h0 as our guess and reiterated the process a couple of times until
the value of h0 does not significantly change.

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216

Chapter 27 Magnetism

  tan1  11cm 1.0cm   28.39  r  3.5cm  7.36 cm
 22cm  3.5cm  sin 28.39
 

 h0  7.36cm  7.36cm2  3.5cm2  0.885cm

  tan1  11 cm  0.885 cm   28.67  r  3.5 cm  7.30 cm
 22 cm  3.5 cm  sin 28.67

 h0  7.30 cm  7.30 cm2  3.5 cm2  0.894 cm

  tan1  11 cm  0.894 cm   28.65  r  3.5 cm  7.30 cm
 22 cm  3.5 cm  sin 28.67

 h0  7.30cm  7.30cm2  3.5cm2  0.894cm

The magnetic field can be determined from the trajectory’s radius, as done in Example 27-7.

   r
 mv  B  mv  m 2eV  2mV  2 9.111031kg 25103V
eB er er m er2
1.60 1019 C 0.0730 m2  7.3mT

72. (a) As the electron orbits the nucleus in the absence of the magnetic field, its

centripetal acceleration is caused solely by the electrical attraction between

the electron and the nucleus. Writing the velocity of the electron as the

circumference of its orbit times its frequency, enables us to obtain an

equation for the frequency of the electron’s orbit.

ke2  m v2  m  2 rf 0 2  f02  ke2
r2 r r 4 2mr3

When the magnetic field is added, the magnetic force adds or subtracts from the centripetal

acceleration (depending on the direction of the field) resulting in the change in frequency.

ke2  q 2 rf B  m 2 rf 2  f 2  qB f  f 2 0
r2 2 m 0
r

We can solve for the frequency shift by setting f  f0  f , and only keeping the lowest order

terms, since f  f0.

 f0  f 2  qB  f0  f  f02  0
2 m

f 2  2 f0f  f 2  qB f0  qB f  f02  0  f   qB
0 2 m 2 m 4 m

(b) The “” indicates whether the magnetic force adds to or subtracts from the centripetal
acceleration. If the magnetic force adds to the centripetal acceleration, the frequency increases.
If the magnetic force is opposite in direction to the acceleration, the frequency decreases.

73. The speed of the proton can be calculated based on the radius of curvature of the (almost) circular
motion. From that the kinetic energy can be calculated.

qvB  mv2  v  qBr K  1 mv2  1 m  qBr 2  q2B2r2
rm 2 2 m  2m

1.60 1019 C 2 0.018T2 
       K  q2B2 2 2
 2m 
r22  r12  2 1.67 1027 kg  8.5 103 m  10.0 103 m

 6.9 1020 J or  0.43eV

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217

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

74. The forces on each of the two horizontal sides of the loop

have the same magnitude, but opposite directions, so these

forces sum to zero. The left side of the loop is located at

x = b, where the magnetic field is zero, and therefore the

force is zero. The net force is the force acting on the right

side of the loop. By the right hand rule, with the current

directed toward the top of the page and the magnetic field

into the page, the force will point in the negative x

direction with magnitude given by Eq. 27-2.

  ˆi  IaB0 1  b  a  ˆi   Ia 2 B0 ˆi
b  b
F  I lB

75. We assume that the horizontal component of the Earth’s magnetic field is  N
pointing due north. The Earth’s magnetic field also has the dip angle of FB
22o. The angle between the magnetic field and the eastward current is
90o. Use Eq. 27-1 to calculate the magnitude of the force. 68o
22o
 F  I lB sin  330 A 5.0 m 5.0 105T sin 90o 
B Earth
E
 0.083 N

Using the right hand rule with the eastward current and the Earth’s magnetic field, the force on the
wire is northerly and 68o above the horizontal .

76. Since the magnetic and gravitational force along the entire rod is
uniform, we consider the two forces acting at the center of mass of the
rod. To be balanced, the net torque about the fulcrum must be zero.
Using the usual sign convention for torques and Eq. 10-10a, we solve
for the magnetic force on the rod.

  0  Mg  1 d   mg  1 d  FM  1 d   FM  M  m g
4 4 4

We solve for the current using Eq. 27-2.

I  F  M mg  8.0m  m g  7.0mg
lB dB
dB dB

The right hand rule indicates that the current must flow toward the left since the magnetic field is
into the page and the magnetic force is downward.

77. (a) For the rod to be in equilibrium, the gravitational torque and the
magnetic torque must be equal and opposite. Since the rod is uniform,
the two torques can be considered to act at the same location (the center
of mass). Therefore, components of the two forces perpendicular to the
rod must be equal and opposite. Since the gravitational force points
downward, its perpendicular component will point down and to the right.
The magnetic force is perpendicular to the rod and must point towards
the left to oppose the perpendicular component of the gravitational force.
By the right hand rule, with a magnetic field pointing out of the page, the

current must flow downward from the pivot to produce this force.

(b) We set the magnitude of the magnetic force, using Eq. 27-2, equal to the magnitude of the
perpendicular component of the gravitational force, F  mg sin , and solve for the magnetic

field.

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218

Chapter 27 Magnetism

 I lB  mg sin
 B  mg sin  0.150 kg 9.80 m/s2 sin13  0.028T
Il 12 A1.0m

(c) The largest magnetic field that could be measured is when   90.

 Bmax
 mg sin 90  0.150 kg 9.80 m/s2 sin 90  0.12 T
Il 12 A1.0m

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219

CHAPTER 28: Sources of Magnetic Field

Responses to Questions

1. Alternating currents will have little effect on the compass needle, due to the rapid change of the

direction of the current and of the magnetic field surrounding it. Direct currents will deflect a
compass needle. The deflection depends on the magnitude and direction of the current and the
distance from the current to the compass. The effect on the compass decreases with increasing
distance from the wire.

2. The magnetic field due to a long straight current is proportional to the current strength. The electric
field due to a long straight line of electric charge at rest is proportional to the charge per unit length.
Both fields are inversely proportional to the distance from the wire or line of charge. The magnetic
field lines form concentric circles around the wire; the electric field lines are directed radially
outward if the line of charge is positive and radially inward if the line of charge is negative.

3. The magnetic forces exerted on one wire by the other try to align the wires. The net force on either
wire is zero, but the net torque is not zero.

4. Yes. Assume the upper wire is fixed in position. Since the currents in the wires are in the same
direction, the wires will attract each other. The lower wire will be held in equilibrium if this force of
attraction (upward) is equal in magnitude to the weight of the wire (downward).

5. (a) The current in the lower wire is opposite in direction to the current in the upper wire.
(b) The upper wire can be held in equilibrium due to the balance between the magnetic force from
the lower wire and the gravitational force. The equilibrium will be stable for small vertical
displacements, but not for horizontal displacements.

6. (a) Let I2  I1. 
(b) Let I2  I1.  B  d l  20I1


  I0 encl  0 I1  I2

B  d l   I0 encl  0  I1  I2   0


7. Inside the cavity B  0 since the geometry is cylindrical and no current is enclosed.

8. Construct a closed path similar to that shown in part (a) of the figure, such that sides ab and cd are
perpendicular to the field lines and sides bc and da lie along the field lines. Unlike part (a), the path
will not form a rectangle; the sides ab and cd will flare outward so that side bc is longer than side da.
Since the field is stronger in the region of da than it is in the region of bc, but da is shorter than bc,

the contributions to the integral in Ampère’s law may cancel. Thus,  I0 encl  B  d   0 is possible

and the field is consistent with Ampère’s law. The lines could not curve upward instead of
downward, because then bc would be shorter than da and it would not be possible for the
contributions to sum to zero.

9. The equation for the magnetic field strength inside a solenoid is given by B  0nI.

(a) The magnetic field strength is not affected if the diameter of the loops doubles.
(b) If the spacing between the loops doubles, the number of loops per unit length decreases by a

factor of 2, and the magnetic field strength then also decreases by a factor of 2.

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220

Chapter 28 Sources of Magnetic Field

(c) If the solenoid’s length is doubled along with the doubling of the total number of loops, then the
number of loops per unit length remains the same, and the magnetic field strength is not
affected.

10. The Biot-Savart law states that the net field at a point in space is the vector sum of the field
contributions due to each infinitesimal current element. As shown in Example 28-12, the magnetic
field along the axis of a current loop is parallel to the axis because the perpendicular field
contributions cancel. However, for points off the axis, the perpendicular contributions will not
cancel. The net field for a point off the axis will be dominated by the current elements closest to it.
For example, in Figure 28-21, the field lines inside the loop but below the axis curve downward,
because these points in space are closer to the lower segment of the loop (where the current goes into
the page) than they are to the upper segment (where the current comes out of the page).

11. No. The magnetic field varies in strength and direction for points in the plane of the loop. The
magnetic field is strongest at the center of the loop.

12. The lead-in wires to electrical devices have currents running in opposite directions. The magnetic
fields due to these currents are therefore also opposite in direction. If the wires are twisted together,
then the distance from a point outside the wires to each of the individual wires is about the same, and
the field contributions from the two wires will cancel. If the wires were not twisted and were
separate from each other, then a point outside the wires would be a different distance from one of the
wires than from the other, and there would be a net field due to the currents in the wires.

13. The Biot-Savart law and Coulomb’s law are both inverse-square in the radius and both contain a

proportionality constant. Coulomb’s law describes a central force; the Biot-Savart law involves a
cross product of vectors and so cannot describe a central force.

14. (a) The force between two identical electric charges is given by Coulomb’s law: F  kq2 .
r2

Magnetic pole strength of a bar magnet could be defined using an analogous expression for the

magnetic force between the poles of two identical magnets: F  m2 . Then, magnetic pole
4 r2

strength, m, would be given by m  4 Fr2 . To determine m, place two identical magnets


with their poles facing each other a distance r apart and measure the force between them.
(b) The magnetic pole strength of a current loop could be defined the same way by using two

identical current loops instead of two bar magnets.

15. Determine the magnetic field of the Earth at one of the magnetic poles (north or south), and use
Equation 28-7b to calculate the magnetic moment. In this equation, x will be (approximately) the
radius of the Earth.

16. To design a relay, place an iron rod inside a solenoid, with the solenoid
solenoid oriented such that one end of it is facing a second iron
rod on a pivot. The second iron rod functions as a switch for iron rod on pivot
the large-current circuit and is normally held open by a spring.
When current flows through the solenoid, the iron rod inside it
becomes magnetized and attracts the second iron rod, closing
the switch and allowing current to flow.

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221

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

17. (a) The source of the kinetic energy is the attractive force produced by the magnetic field from the
magnet acting on the magnetic moments of the atoms in the iron.

(b) When the block strikes the magnet, some of the kinetic energy from the block is converted into
kinetic energy in the iron atoms in the magnet, randomizing their magnetic moments and
decreasing the overall field produced by the magnet. Some of the kinetic energy of the block as
a whole is also converted into the kinetic energy of the individual atoms in the block, resulting
in an increase in thermal energy.

18. No, a magnet with a steady field will only attract objects made of ferromagnetic materials.
Aluminum is not ferromagnetic, so the magnetic field of the magnet will not cause the aluminum to
become a temporary magnet and therefore there will be no attractive force. Iron is ferromagnetic, so
in the presence of a magnet, the domains in a piece of iron will align such that it will be attracted to
the magnet.

19. An unmagnetized nail has randomly oriented domains and will not generate an external magnetic
field. Therefore, it will not attract an unmagnitized paper clip, which also has randomly oriented
domains. When one end of the nail is in contact with a magnet, some of the domains in the nail align,
producing an external magnetic field and turning the nail into a magnet. The magnetic nail will cause
some of the domains in the paper clip to align, and it will be attracted to the nail.

20. Yes, an iron rod can attract a magnet and a magnet can attract an iron rod. Consider Newton’s third
law. If object A attracts object B then object B attracts object A.

21. Domains in ferromagnetic materials in molten form were aligned by the Earth’s magnetic field and
then fixed in place as the material cooled.

22. Yes. When a magnet is brought near an unmagnetized piece of iron, the magnet’s field causes a
temporary alignment of the domains of the iron. If the magnet’s north pole is brought near the iron,
then the domains align such that the temporary south pole of the iron is facing the magnet, and if the
magnet’s south pole is closest to the iron, then the alignment will be the opposite. In either case, the
magnet and the iron will attract each other.

23. The two rods that have ends that repel each other will be the magnets. The unmagnetized rod will be
attracted to both ends of the magnetized rods.

24. No. If they were both magnets, then they would repel one another when they were placed with like
poles facing each other. However, if one is a magnet and the other isn’t, they will attract each other
no matter which ends are placed together. The magnet will cause an alignment of the domains of the
non-magnet, causing an attraction.

25. (a) The magnetization curve for a paramagnetic substance is a straight line with slope slightly
greater than 1. It passes through the origin; there is no hysteresis.

(b) The magnetization curve for a diamagnetic substance is a straight line with slope slightly less
than 1. It passes through the origin; there is no hysteresis.

The magnetization curve for a ferromagnetic substance is a hysteresis curve (see Figure 28-29).

26. (a) Yes. Diamagnetism is present in all materials but in materials that are also paramagnetic or
ferromagnetic, its effects will not be noticeable.

(b) No. Paramagnetic materials are nonferromagnetic materials with a relative permeability greater
than one.

(c) No. Ferromagnetic materials are those that can be magnetized by alignment of their domains.

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222

Chapter 28 Sources of Magnetic Field

Solutions to Problems

1. We assume the jumper cable is a long straight wire, and use Eq. 28-1.

 Bcable
 0 I  4 107 Tm A 65A  3.714 104 T  3.7 104 T
2 r 2 0.035m

Compare this to the Earth’s field of 0.5 104 T .

Bcable BEarth  3.714 104 T  7.43 , so the field of the cable is over 7 times that of the Earth.
5.0 105T

2. We assume that the wire is long and straight, and use Eq. 28-1.

 Bwire
 0 I  I  2 rBwire  2 0.15m 0.50 104 T  37.5A  38 A
2 r 0
4 107 Tm A

3. Since the currents are parallel, the force on each wire will be attractive, toward the other wire. Use

Eq. 28-2 to calculate the magnitude of the force.

 F2
 0 I1I2 l2  4 107 Tm A   35 A2   25 m  0.15 N, attractive
2 d 2
0.040 m

4. Since the force is attractive, the currents must be in the same direction, so the current in the second

wire must also be upward. Use Eq. 28-2 to calculate the magnitude of the second current.

F2  0 I1I2 l2 
2 d

2 F2 d 2 0.070 m
0 l2 I1  4 107 Tm A 28 A
 I2
 7.8 104 N m  9.75A  9.8A upward

5. To find the direction, draw a radius line from the wire to the field point. C D
Then at the field point, draw a perpendicular to the radius line, directed so
that the perpendicular line would be part of a counterclockwise circle.

I
E

6. For the experiment to be accurate to 2.0% , the magnetic field due to the current in the cable must

be less than or equal to 2.0% of the Earth’s magnetic field. Use Eq. 28-1 to calculate the magnetic
field due to the current in the cable.

 Bcable
 0 I  0.020BEarth  I  2 r 0.020BEarth   2 1.00 m 0.020 0.5 104 T  5.0 A
2 r
0 4 107 Tm A

Thus the maximum allowable current is 5.0 A .

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7. Since the magnetic field from a current carrying wire circles the

wire, the individual field at point P from each wire is perpendicular
to the radial line from that wire to point P. We define B1 as the


field from the top wire, and B2 as the field from the bottom wire.

We use Eq. 28-1 to calculate the magnitude of each individual

field.

 B1
 0 I  4 107 Tm A 35 A  1.17  104 T
2 r1 2 0.060 m

 B2
 0 I  4 107 Tm A 35 A  7.00  105 T
2 r2 2 0.100 m

We use the law of cosines to determine the angle that the radial

line from each wire to point P makes with the vertical. Since the field is perpendicular to the radial

line, this is the same angle that the magnetic fields make with the horizontal.

1  cos1  0.060 m2  0.130 m2  0.100 m2   47.7
 20.060 m0.130 m 

2  cos1  0.100 m2  0.130 m2  0.060 m2   26.3
 20.100 m0.130 m 

Using the magnitudes and angles of each magnetic field we calculate the horizontal and vertical

components, add the vectors, and calculate the resultant magnetic field and angle.

    Bnet x  B1 cos 1  B2 cos2  1.174 104 T cos 47.7o  7.00 105 T cos 26.3o  1.626 105 T

    Bnet y  B1 sin 1  B2 sin1  1.17 104 T sin 47.7o  7.00 105 T sin 26.3o  1.18 104 T

   B  2 2 2  1.19 104
B  B2 net, y  1.626 105 T 1.18 104 T T
net , x 

 =tan1 Bnet, y  tan 1 1.18 104 T  82.2
Bnet, y 1.626 105 T


B  1.19 104 T @ 82.2  1.2 104 T @ 82

8. At the location of the compass, the magnetic field caused by the wire will point to  
Bnet BEarth
the west, and the Earth’s magnetic field points due North. The compass needle

will point in the direction of the NET magnetic field. Bwire

 Bwire
 0 I  4 107 Tm A 43A  4.78 105T
2 r 2 0.18m

  tan1 BEarth  tan1 4.5 105T  43N of W
Bwire 4.78 105T

9. The magnetic field due to the long horizontal wire points straight up at the 
Bwire
point in question, and its magnitude is given by Eq. 28-1. The two fields are

oriented as shown in the diagram. The net field is the vector sum of the two

fields.  
 Bnet
 Bwire 107 BEarth
 0 I  4 2 Tm A 24.0 A  2.40  105 T
2 r  0.200 m

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224

Chapter 28 Sources of Magnetic Field

BEarth  5.0 105T Bnet y  Bwire  BEarth sin 44o  1.07 105T
Bnet x  BEarth cos 44o  3.60 105 T

   Bnet  2 2
B  B2 2  3.60 105T 1.07 105T 3.8 105T
net x net y  

  tan1 Bnet y  tan1 1.07 105T  17 below the horizontal
Bnet x 3.60 105T

10. The stream of protons constitutes a current, whose magnitude is found by multiplying the proton rate

times the charge of a proton. Then use Eq. 28-1 to calculate the magnetic field.

    Bstream
 0 I  4 107 Tm A 2.5 109 protons s 1.60 1019 C proton  4.0 1017 T
2 r
2 2.0m

11. (a) If the currents are in the same direction, the magnetic fields at the midpoint between the two

currents will oppose each other, and so their magnitudes should be subtracted.

   Bnet
 0 I1  0 I 2  4 107 Tm A  I  25A  2.0 105 T A  I  25A
2 r1 2 r2
2 0.010 m

(b) If the currents are in the opposite direction, the magnetic fields at the midpoint between the two

currents will reinforce each other, and so their magnitudes should be added.

   Bnet
 0 I1  0 I 2  4 107 Tm A  I  25A  2.0 105 T A  I  25A
2 r1 2 r2
2 0.010 m

12. Using the right-hand-rule we see that if the currents flow in the same direction, the magnetic fields
will oppose each other between the wires, and therefore can equal zero at a given point. Set the sum
of the magnetic fields from the two wires equal to zero at the point 2.2 cm from the first wire and use
Eq. 28-1 to solve for the unknown current.

Bnet  0  0 I1  0 I 2  I2   r2  I1   6.0cm  2.2 cm  2.0 A  3.5 A
2 r1 2 r2  r1   2.2 cm 
   

13. Use the right hand rule to determine the direction of the magnetic field

from each wire. Remembering that the magnetic field is inversely
proportional to the distance from the wire, qualitatively add the magnetic
field vectors. The magnetic field at point #2 is zero.

14. The fields created by the two wires will oppose each other, so the net field is the difference of the
magnitudes of the two fields. The positive direction for the fields is taken to be into the paper, and
so the closer wire creates a field in the positive direction, and the farther wire creates a field in the
negative direction. Let d be the separation distance of the wires.

Bnet  0 I  0I  0 I 1 1  0 I 1  1 
2 rcloser 2 rfarther 2  rcloser   2  
  r  1 d r  1 d 
rfarther 2 2

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 0 I  r  1 d d  1 d 
2  2 2 
 r 

  4 107 Tm A 28.0 A  0.0028 m 
 
2  0.10 m  0.0014 m 0.10 m  0.0014 m 

 1.568 106 T  1.6 106 T

Compare this to the Earth’s field of 0.5 104 T .

Bnet BEarth  1.568 106T  0.031
0.5 104 T

The field of the wires is about 3% that of the Earth.

15. The center of the third wire is 5.6 mm from the left wire, and 2.8 mm from the right wire. The force
on the near (right) wire will attract the near wire, since the currents are in the same direction. The
force on the far (left) wire will repel the far wire, since the currents oppose each other. Use Eq. 28-2
to calculate the force per unit length.

Fnear  0 I1I2 l near 
2 d near

Fnear  0 I1I2  4 107 Tm A 25.0 A 28.0 A  0.050 N m, attractive
l near 2 d near 2
2.8 103 m

Ffar  0 I1I2 lfar 
2 dfar

Ffar  0 I1I2  4 107 Tm A 25.0A 28.0 A  0.025 N m, repelling
lfar 2 dfar 2
5.6 103 m

16. (a) We assume that the power line is long and straight, and use Eq. 28-1.

 Bline
 0 I  4 107 Tm A  95 A   2.235106 T  2.2106 T
2 r
2 8.5m

The direction at the ground, from the right hand rule, is south. Compare this to the Earth’s field

of 0.5 104 T , which points approximately north.

Bline BEarth  2.235106 T  0.0447
0.5104 T

The field of the cable is about 4% that of the Earth.

(b) We solve for the distance where Bline  BEarth.

   Bline
 0 I  BEarth  r  0I  4 107 Tm A 95A  0.38m  0.4 m
2 r 2 BEarth
2 0.5 104 T

So about 0.4 m below the wire, the net B-field would be 0, assuming the Earth’s field points
straight north at this location.

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226

Chapter 28 Sources of Magnetic Field

17. The Earth’s magnetic field is present at both locations in the problem, and we

assume it is the same at both locations. The field east of a vertical wire must be 
BEarth 
pointing either due north or due south. The compass shows the direction of the net 
magnetic field, and it changes from 28o E of N to 55o E of N when taken inside. Bwire

That is a “southerly” change (rather than a “northerly” change), and so the field 
due to the wire must be pointing due south. See the diagram. For the angles,

  28,     55, and       180 and so   27 and   125o . Use  Bnet
the law of sines to find the magnitude of Bwire , and then use Eq. 28-1 to find the
magnitude of the current.

B  Bwire Earth  Bwire  BEarth sin  0 I 
sin sin sin 2 r

 I sin 2 sin 27o 2
 BEarth sin 0 r  5.0 105T sin125o 4 107 Tm A 0.120m  17 A

Since the field due to the wire is due south, the current in the wire must be down .

18. The magnetic field at the loop due to the long wire is into the page, and can be calculated by Eq. 28-
1. The force on the segment of the loop closest to the wire is towards the wire, since the currents are
in the same direction. The force on the segment of the loop farthest from the wire is away from the
wire, since the currents are in the opposite direction.

Because the magnetic field varies with distance, it is more difficult to calculate the total force on the
left and right segments of the loop. Using the right hand rule, the force on each small piece of the
left segment of wire is to the left, and the force on each small piece of the right segment of wire is to
the right. If left and right small pieces are chosen that are equidistant from the long wire, the net
force on those two small pieces is zero. Thus the total force on the left and right segments of wire is
zero, and so only the parallel segments need to be considered in the calculation. Use Eq. 28-2.

Fnet  Fnear  Ffar  0 I1I2 l near  0 lI1I2  0 I1I 2 l  1 1
2 d near 2 d far 2  d near  dfar 

far

 4 107 Tm A  3.5 A2  0.100 m  1  1   5.1106 N, towards wire
2  0.030 0.080 
m m

19. The left wire will cause a field on the x axis that points in the y direction, and the right wire will

cause a field on the x axis that points in the negative y direction. The distance from the left wire to a
point on the x axis is x, and the distance from the right wire is d  x .

  0 I ˆj  2 0 I x ˆj  0 I  1  d 1 x  ˆj  0 I  d  2x  ˆj
Bnet 2 x 2  x   2  
d   xd  x 

20. The left wire will cause a field on the x axis that points in the negative y direction, and the right wire
will also cause a field on the x axis that points in the negative y direction. The distance from the left
wire to a point on the x axis is x, and the distance from the right wire is d  x .

   0 2I  ˆj  0 I ˆj   0I  2  1  ˆj
Bnet  2  x  
2 x 2 d x  d x

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21. The magnetic fields created by the individual currents will be at right angles to each other. The field
due to the top wire will be to the right, and the field due to the bottom wire will be out of the page.
Since they are at right angles, the net field is the hypotenuse of the two individual fields.

  I0 top 2   I0 bottom 2 0 4 107 Tm A 20.0 A2  12.0A2
 2 rtop   2 rbottom  2 r 2 0.100 m
   I 2
top
Bnet     I2 
bottom

 4.66 105T

22. The net magnetic field is the vector sum of the magnetic fields

produced by each current carrying wire. Since the individual

magnetic fields encircle the wire producing it, the field is

perpendicular to the radial line from the wire to point P. We let
B1 be the field from the left wire, and B2 designate the field
from the right wire. The magnitude of the magnetic field

vectors is calculated from Eq. 28-1.

 B1
 0 I  4 107 Tm A 16.5 A  2.7500 105 T
2 r1 2 0.12 m

 B2
 0 I  4 107 Tm A 16.5A  2.5385 105 T
2 r2 2 0.13m

We use the law of cosines to determine the angle that the radial line from each wire to point P makes

with the horizontal. Since the magnetic fields are perpendicular to the radial lines, these angles are

the same as the angles the magnetic fields make with the vertical.

1  cos1  0.12 m2  0.082 m2  0.13 m2   77.606
 20.12 m0.082 m 

2  cos1  0.13 m2  0.082 m2  0.12 m2   64.364
 20.13 m0.082 m 

Using the magnitudes and angles of each magnetic field we calculate the horizontal and vertical

components, add the vectors, and calculate the resultant magnetic field and angle.

   Bnet x  B1 sin 1   B2 sin2   2.7500 105T sin 77.606  2.5385 105T sin 64.364

 49.75 106 T

   Bnet y  B1 cos1   B2 cos1  2.7500 105T cos77.606  2.5385 105T cos64.364

 5.080 106T

   B  2 2
B  B2 2  49.75 106 T 5.080 106 T  5.00 105 T
net , x net, y 

 =tan 1 Bnet, y  tan1 5.08 106 T  5.83
Bnet, y 49.75 106 T


B  5.00 105 T @ 5.83 below the negative x-axis

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228

Chapter 28 Sources of Magnetic Field

23. (a) The net magnetic field a point y above the center of

the strip can be found by dividing the strip into
infinitely thin wires and integrating the field
contribution from each wire. Since the point is directly

above the center of the strip, we see that the vertical
contributions to the magnetic field from symmetric

points on either side of the center cancel out. Therefore,
we only need to integrate the horizontal component of

the magnetic field. We use Eq. 28-1 for the magnitude
of the magnetic field, with the current given by

dI  I dx.
d

0 sin dI  0I 1d dx  y  0 Iy 1d dx
x2  y2   2 d x2  y2
2 x2  2

 Bx  
2 r 2 d  y2
1 d  1 d
2 2

 0 Iy 1 tan1 x d / 2  0 I tan 1  d 
2 d y y d /2 d  2y 

(b) In the limit of large y, tan1 d / 2 y  d / 2 y.

Bx  0 I tan 1  d   0 I d  0 I
d  2y  d 2y 2 y
 

This is the same as the magnetic field for a long wire.

24. We break the current loop into the three branches of the triangle and add the forces from each of the

three branches. The current in the parallel branch flows in the same direction as the long straight

wire, so the force is attractive with magnitude given by Eq. 28-2.

F1  0II  a
2 d

By symmetry the magnetic force for the other two segments will be equal. These two wires can be

broken down into infinitesimal segments, each with horizontal length dx. The net force is found by

integrating Eq. 28-2 over the side of the triangle. We set x=0 at the left end of the left leg. The

distance of a line segment to the wire is then given by r  d  3x . Since the current in these

segments flows opposite the direction of the current in the long wire, the force will be repulsive.

a/2 0II  dx  0II  ln a/2 0II  1  3a 
d  3x 2 3 0 2 3 2d 
    F2 0 2 d 3x  ln

We calculate the net force by summing the forces from the three segments.

F  F1  2F2  0II  a  2 0II  ln 1  3a   0II  a  3 ln 1  3a 
2 d 2 3 2d    3 2d 
 2d

25. Use Eq. 28-4 for the field inside a solenoid.

 B  0IN
 l
 I  Bl  0.385 103 T 0.400 m
0 N 4 107 Tm A 765  0.160 A

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229

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

26. The field inside a solenoid is given by Eq. 28-4.

N  Bl  0.30T 0.32 m
0 I 107 Tm A 4.5A
 B  0IN  4  1.7 104 turns
l

27. (a) We use Eq. 28-1, with r equal to the radius of the wire.

   B
 0 I  4 107 Tm/A 33 A  5.3 mT
2 r
2 1.25 103 m

(b) We use the results of Example 28-6, for points inside the wire. Note that r  1.25  0.50 mm

 0.75mm.

     B
 0 Ir  4 107 Tm/A 33A 0.75 103 m  3.2 mT
2 R2
2 1.25 103 m 2

(c) We use Eq. 28-1, with r equal to the distance from the center of the wire.

  B
 0 I  2 4 107 Tm/A 33 A  1.8 mT
2 r
1.25 103 m  2.5 103 m

28. We use the results of Example 28-10 to find the maximum and minimum fields.

 Bmin
 0 NI  4 107 Tm/A 687 25.0 A  12.7 mT
2 rmax 2 0.270 m

 Bmax
 0 NI  4 107 Tm/A 687 25.0 A  13.7 mT
2 rmin 2 0.250 m

12.7 mT  B  13.7 mT

29. (a) The copper wire is being wound about an average Dd d
diameter that is approximately equal to the outside D
diameter of the solenoid minus the diameter of the wire, or
D  d. See the (not to scale) end-view diagram. The

length of each wrapping is   D  d . We divide the

length of the wire L by the length of a single winding to
determine the number of loops. The length of the solenoid
is the number of loops multiplied by the outer diameter of
the wire, d.

L 20.0 m
   l D
 d   d   2.00 103 m  2.50 102 m  2.00 103 m   0.554 m

(b) The field inside the solenoid is found using Eq. 28-4. Since the coils are wound closely

together, the number of turns per unit length is equal to the reciprocal of the wire diameter.

L

n  # turns   D  d  ld  1
l l d
l

 B
 0nI  0 I  4 107 Tm/A 16.7 A  10.5 mT
d
2.00 103 m

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230

Chapter 28 Sources of Magnetic Field

30. (a) The magnitude of the magnetic field from each wire is found using Eq. 28-1. The direction of

the magnetic field is perpendicular to the radial vector from the current to the point of interest.

Since the currents are both coming out of the page, the magnetic fields will point

counterclockwise from the radial line. The total magnetic field is the vector sum of the

individual fields.
  
B1 B2 0 I 0 I
   B 2 r1 2 r2
    sin1ˆi  cos1ˆj   sin2ˆi  cos2ˆj

 0 I   sin1  sin1  ˆi   cos1  cos2  ˆj
2  r1 r2   r1 r2 
   

This equation for the magnetic field shows

that the x-component of the magnetic field

is symmetric and the y-component is anti-

symmetric about  = 90.

(b) See sketch.

(c) The two diagrams are similar in shape, as

both form loops around the central axes.

However, the magnetic field lines form a

vector field, showing the direction, not

necessarily the magnitude of the magnetic

field. The equipotential lines are from a

scalar field showing the points of constant

magnitude. The equipotential lines do not

have an associated direction.

31. Because of the cylindrical symmetry, the magnetic fields will be

circular. In each case, we can determine the magnetic field using

Ampere’s law with concentric loops. The current densities in the wires R1
r R2
are given by the total current divided by the cross-sectional area.
I0 (out) R3
I0 I0
 R12 R32  R22 I0 (in)
J inner   J outer 


(a) Inside the inner wire the enclosed current is determined by the

Bde ndssity of the inner wire.
  current 0 Iencl  0 J inner R2

B 2 R  0 I0 R2  B 0 I 0 R
 R12 2 R12

(b) Between the wires the current enclosed is the current on the inner wire.

   ds  0 Iencl  B 2 R  0I0  B  0 I 0
B 2 R

(c) Inside the outer wire the current enclosed is the current from the inner wire and a portion of the

the outer wire.
 0Iencl  0  I0
  currentBfrodms
 J outer R2  R22 

    
2 r I0 R2  R22  0 I 0 R32  R2
R32  R22  2 R R32  R22
   B  0  I0    B


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231

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

(d)OutsideBth desouter0 wire the net current enclosed is zero.
Iencl  0  B0
B 2 R  0 

(e) See the adjacent graph. The B (10-5 T) 3.0
spreadsheet used for this problem 2.5
can be found on the Media
Manager, with filename 2.0
“PSE4_ISM_CH28.XLS,” on tab 1.5
“Problem 28.31e.” 1.0

0.5
0.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0

R (cm)

32. We first find the constants C1 and C2 by integrating the currents over each cylinder and setting the
integral equal to the total current.

R1 RdR R1 R2dR 2 R13C1 C1 3I0
3 2 R13
C1R2 2 C1

0 0
 I0     

R3 2 3I0
R2 3 R33  R23
    I0 R2dR
 2 C2   R33  R23 C2  C2  2

(a) Inside the inner wire the enclosed current is determined by integrating the current density inside

the radius R.

  ds R C1R 2 2 0 C1R3 2  3I0  R3
B 0 3 3  2 R13 
  0 Iencl  0 RdR   0  

B 2 R  0 I0 R3  B 0 I 0 R 2
 R13 2 R13

(b) Between the wires the current enclosed is the current on the inner wire.

   ds  0 Iencl  B 2 R  0I0  B  0 I 0
B 2 R

(c) Inside the outer wire the current enclosed is the current from the inner wire and a portion of the

current from the outer wire.
   ds
R  R 
B R2  R2 
  0 Iencl  0  I 0  C2 R 2 RdR  0 I0   C2 R 2 RdR


    1  2   R3  R23 
0 I 0 3  C2 R3  R23   0  I0  R33  R23 
 

      B 2 r R33  R23 R3  R23 R33  R3
 R33  R23  R33  R23  B 0 I 0 R33  R23
 0I0   2 R



(d)OutsideBth desouter0 wire the net current enclosed is zero.
Iencl  0  B0
B 2 R  0 

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232

Chapter 28 Sources of Magnetic Field

33. Use Eq. 28-7b to write a ratio of the magnetic fields at the surface of the earth and 13,000 km above

the surface. Use the resulting ratio to determine the magnetic field above the surface.

0   6.38 103km 3

B2 2 x23 x13   x13 = 3.6 106T
B1  0   x23 B2 =B1 x23 = 1.0 104T  19.38 103km 
 
2 x13

34. Since the point C is along the line of the two straight segments of the current, I

these segments do not contribute to the magnetic field at C. We calculate the R2
R1
magnetic field by integrating Eq. 28-5 along the two curved segments. Along I

each integration the line segment is perpendicular to the radial vector and the

   radialBdista4n0IceR01isdclRo12nsrˆtant4.0I0   0I kˆ R1 ds  0 I 0 ds C
R2 d l  rˆ 4 R12 4 R22
0 kˆ
R22
R2

 0I kˆ  0 I kˆ  0 I  R2  R1  kˆ
4 R1 4 R2 4  R1R2 
 

35. Since the current in the two straight segments flows radially toward and away from the center of the

loop, they do not contribute to the magnetic field at the center. We calculate the magnetic field by

integrating Eq. 28-5 along the two curved segments. Along each integration segment, the current is

perpendicular to the radial vector and the radial distance is constant. By the right-hand-rule the

magnetic field from the upper portion will point into the page and the magnetic field from the lower

portion will point out of the page.

   0 I1 ds kˆ  0 I 2 ds kˆ  0  R  kˆ  I1  I   0 kˆ  0.35I  0.65I    30 I
4 R2 4 R2 4R 40R
B 4 R2 2

upper lower

36. We assume that the inner loop is sufficiently small that the magnetic field from the

larger loop can be considered to be constant across the surface of the smaller loop. μ 2 I1

The field at the center of the larger loop is illustrated in Example 28-12. Use Eq. 27- B1

10 to calculate the magnetic moment of the small loop, and Eq. 27-11 to calculate the I2
R1
torque. 
 IA
B  0I ˆi μ   I R22ˆj
2R

τ  μ    I R22ˆj  0I ˆi   0 I 2R22 kˆ
B 2R 2R

Tm/A  7.0 A2 0.018 m2
20.25 m
  
4 107 kˆ  1.3107 kˆ mN

This torque would cause the inner loop to rotate into the same plane as the outer loop with the
currents flowing in the same direction.

37. (a) The magnetic field at point C can be obtained using the Biot- I R1
C
Savart law (Eq. 28-5, integrated over the current). First break the
loop into four sections: 1) the upper semi-circle, 2) the lower R2
semi-circle, 3) the right straight segment, and 4) the left straight I
segment. The two straight segments do not contribute to the
magnetic field as the point C is in the same direction that the

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233

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

current is flowing. Therefore, along these segments rˆ and dˆ are parallel and dˆ  rˆ  0 . For

the upper segment, each infinitesimal line segment is perpendicular to the constant magnitude

radial vector, so the magnetic field points downward with constant magnitude.

 0 I dˆ  rˆ  0I kˆ  R1    0 I kˆ .
4 r2 4 R12 4R1
Bupper 

Along the lower segment, each infinitesimal line segment is also perpendicular to the constant

radial vector.

 0 I dˆ  rˆ  0 I kˆ  R2    0 I kˆ
4 r2 4 R22 4R2
Blower 

Adding the two contributions yields the total magnetic field.

    0 I kˆ  0 I kˆ   0 I 1  1  kˆ
B  Bupper  Blower 4R1 4R2 4  R2 
 R1 

(b) The magnetic moment is the product of the area and the current. The area is the sum of the two
half circles. By the right-hand-rule, curling your fingers in the direction of the current, the
thumb points into the page, so the magnetic moment is in the kˆ direction.

 μ   R12  R22  Ikˆ  I kˆ
   2  2   2 R12  R22
 

38. Treat the moving point charge as a small current segment. We can write the rP
v
product of the charge and velocity as the product of a current and current segment.

InsertBqinvgt4hq0eIddsetldilrn2tdodrˆqtthdel4B0ioIqdtv-rSl2arvˆ art law gives us the magnetic field at point P.

0 qv  r q
4 r3

39. (a) The disk can be broken down into a series of infinitesimal thick rings.
As the charge in each of these rings rotates it produces a current of

magnitude dI   2  dq , where dq is the surface charge density dr
r
multiplied by the area of the ring. We use Eq. 27-10 to calculate the
Q
magnetic dipole moment of each current loop and integrate the dipole

moments to obtain the total magnetic dipole moment. x

 dμ dIA  Q   Q r3drˆi
  R 2 2 rdr 2   r2  R2

μ R Q r 3dr ˆi  QR2 ˆi R
0 R2 4

(b) To find the magnetic field a distance x along the axis of the disk, we again consider the disk as a

series of concentric currents. We use the results of Example 28-12 to determine the magnetic

field from each current loop in the disk, and then integrate to obtain the total magnetic field.
   
0 r 2 0 r 2 Q rdr
dB  r2  x2 r2  x2  R2
2
3 dI  3
22 2

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234

Chapter 28 Sources of Magnetic Field

   0Q ˆi R r3 3 dr  0Q ˆi r2  2x2 R 0Q ˆi  R2  2x2 
  B2 R2 0 r2  x2 2 2 R2 r2  x2 2 R2  R2  x2  2x
 

0

(c) When we take the limit x  R our equation reduces to Eq. 28-7b. μ
x3
  0Q ˆi 2x 1  R2  1  R2  3R4  ...   2x   0Qˆi  R4   0
B 2 R2  2x2  2x2 8x4   2 R2  4x3  2
 
  

40. (a) Choose the y axis along the wire and the x axis passing from the

center of the wire through the point P. With this definition we

calculate the magnetic field at P by integrating Eq. 28-5 over the

length of the wire. The origin is at the center of the wire.
 
l r dyˆj Rˆi  yˆj
R2  y2 3/2
  0 I d l  rˆ 0 I d  0 I 1 d
4 r2 4 4 2
    B 
 r3 

 1 d
2

0 IR 1 d dy
4 2 R2  y2 3/2
kˆ
  

 1 d
2

d /2

 0 IR kˆ y 0 I d 1/ 2 kˆ
4 R2  y2 1/2 2 R 4R2  d2
  R2  

d /2

(b) If we take the limit as d   , this equation reduces to Eq. 28-1.

 0 I d  0 I
 2 R  d2  2 R
 B lim  1/ 2  
4R2
d 

 

41. (a) The magnetic field at point Q can be obtained by integrating Eq. 28-5 over

the length of the wire. In this case, each infinitesimal current segment dl is

parallel to the x axis, as is each radial vector. Since the magnetic field is

proportional to the cross-product of the current segment and the radial

vector, each segment contributes zero field. Thus the magnetic field at point

Q is zero.

(b) The magnetic field at point P is found by integrating Eq. 28-5 over the

length of the current segment.
 
l r dxˆi  xˆi  yˆj
x2  y2 3/2
  d l  rˆ 0 I 0 0
r2 4 d
       B  kˆ
0 I  d   0 I  0 Iy dx
4 4 4 d x2  y2 3/2
r3

0

0 Iy kˆ x 0 I d 1/ 2 kˆ
4 x2  y2 1/2 4 y y2  d2
  y2  
d

42. We treat the loop as consisting of 5 segments, The first has length d,
is located a distance d to the left of point P, and has current flowing
toward the right. The second has length d, is located a distance 2d to
left of point P, and has current flowing upward. The third has length
d, is located a distance d to the left of point P, and has current
flowing downward. The fourth has length 2d, is located a distance d

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235

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

below point P, and has current flowing toward the left. Note that the fourth segment is twice as long

as the actual fourth current. We therefore add a fifth line segment of length d, located a distance d

below point P with current flowing to the right. This fifth current segment cancels the added portion,

but allows us to use the results of Problem 41 in solving this problem. Note that the first line points

radially toward point P, and therefore by Problem 41(a) does not contribute to the net magnetic field.

We add the contributions from the other four segments, with the contribution in the positive z-

direction if the current in the segment appears to flows counterclockwise around the point P.
B  B2  B3  B4  B5

 0 I d kˆ  0 I d kˆ 0 I 2d kˆ 0 I d 1/ 2 kˆ
 d2 4 d  d2 4 d  4d 2 4 d  d2
4 2d
         4d 2 1/ 2 d2 1/ 2  d2 1/ 2  d2

 0 I  2 5  kˆ
4 d  2

43. (a) The angle subtended by one side of a polygon, , from the I0

center point P is 2 divided by the number of sides, n. The P
length of the side L and the distance from the point to the
center of the side, D, are obtained from trigonometric relations.

L  2R sin  / 2  2R sin  / n

D  R cos / 2  R cos / n R
D
The magnetic field contribution from each side can be found
using the result of Problem 40. L

0 I L 0 I 2R sin  / n
2 D L2  4D2
2 R cos 2R sin  / n2  4R cos / n2
   B 1  / n 1
2 2

 0 I tan  / n 
2 R

The contributions from each segment add, so the total magnetic field is n times the field from

one side.

Btotal  0 In tan  /n
2 R

(b) In the limit of large n,  / n , becomes very small, so tan( / n)   / n .

Btotal  0 In   0 I
2 R n 2R

This is the magnetic field at the center of a circle.

44. The equation derived in Eq. 28-12 gives the magnetic field

a distance x from a single loop. We expand this single loop

to the field of an infinite solenoid by multiplying the field

from a single loop by n dx, the density of loops times the

infinitesimal thickness, and integrating over all values of x.

Use the table in Appendix B-4 to evaluate the integral.



 0 IR 2ndx 0IR2n  dx 0 IR 2n x  0In
2  R2  x2 2 R2  x2
B 3
        2 R2  x2 2  3  R2 1
2 2



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236

Chapter 28 Sources of Magnetic Field

45. To find the magnetic field at point (x,y) we break each current segment

into two segments and sum fields from each of the eight segments to

determine the magnetic field at the center. We use the results of Problem

41(b) to calculate the magnetic field of each segment.
   
0 I
B 4 y
 x 1/ 2 kˆ  0 I y2 (b  x) 1/ 2 kˆ
y2  x2 4 y  (b  x)2

0 I y kˆ  0 I (a  y) kˆ
(b  (b  x)2  (b  y)2  (b  x)2
   4 x) y2 1/ 2 4 x) 1/ 2
(a 

0 I (b  x) kˆ  0 I x) 1/ 2 kˆ
(a  (a  y)2  (b  x)2 (a  (a  y)2  x2
   4 y) 1/ 2 4 y)

0 I (a  y) kˆ  0 I y 1/ 2 kˆ
4 x (a  y)2  x2 4 x y2  x2
    1/ 2

We simplify this equation by factoring out common constants and combining terms with similar
roots.

 0 I  y2  x2  y2  (b  x)2  (a  y)2  (b  x)2  (a  y)2  x2  kˆ
B 4  xy (b  x) y (a  y)(b  x) x(a  y) 

46. (a) By symmetry we see that on the x axis the magnetic field can only have an x component. To

justify this assertion, imagine that the magnetic field had a component off the axis. If the

current loop were rotated by 90 about the x axis, the loop orientation would be identical to the

original loop, but the off-axis magnetic field component would have changed. This is not

possible, so the field only has an x component. The contribution to this field is the same for

each loop segment, and so the total magnetic field is equal to 4 times the x component of the

magnetic field from one segment. We integrate Eq. 28-5 to find this magnetic field.

dy ˆj kˆ
  1 d 0 I 1 d 0 Id ˆi 1 d dy
2 4 2 2 2
     B
 4 2 x2 2 3 / 2  2 x2 y2 3/ 2

 1 d  1 d   y   1 d  1 d   
2  2  2  2 

1 d
2

0 Id ˆi y 2 2d 20I ˆi
2 d2  4x2 d2  2x2
      2 x2 2 x2 y2 1/ 2  1/ 2

 1 d    1 d    
 2   2 
 1 d
2

(b) Let x  d to show that the magnetic field reduces to a dipole field of Eq. 28-7b.

   ˆi d 20I ˆi
2 x3
B 1/ 2
 2 2d 20I 
4x2 2x2


Comparing our magnetic field to Eq. 28-7b we see that it is a dipole field with the magnetic
moment μ  d 2I ˆi

47. (a) If the iron bar is completely magnetized, all of the dipoles are aligned. The total dipole moment

is equal to the number of atoms times the dipole moment of a single atom.

  N 1  N AV 1
Mm

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237

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

  
6.022  1023 atoms/mole 7.80g/cm3 9.0cm1.2 cm  1.0 cm  1.8 1023 Am2 
 atom 
55.845g/mole 

 16.35 Am2  16 Am2

(b) We use Eq. 27-9 to find the torque.

   B sin  16.35 Am2 0.80 Tsin 90  13 mN

48. The magnetic permeability is found 70
60
from the two fields. 50
40
B0  0nI ; B  nI ;  (10-4 T-m/A) 30
20
B B 10
    0 B0 0

B0 0 0 2 4 6 8 10 12 14

For the graph, we have not plotted B 0(10-4 T)

the last three data points so that the

structure for low fields is seen. The

spreadsheet used for this problem can

be found on the Media Manager, with

filename “PSE4_ISM_CH28.XLS,”

on tab “Problem 28.48.”

49. The magnetic field of a long, thin torus is the same as the field given by a long solenoid, as in Eq.
28-9.

  B  nI  2200 4 107 Tm/A 285 m-1 3.0 A  2.4 T

50. The field inside the solenoid is given by Eq. 28-4 with 0 replaced by the permeability of the iron.

B  NI    Bl  2.2 T 0.38 m  2.7 105 Tm A  220
l NI 640 48 A

51. Since the wires all carry the same current and are

equidistant from each other, the magnitude of the force per

unit length between any two wires is the same and is given

by Eq. 28-2.

4 107 Tm/A 8.00 A2
2 0.035 m
 F0I 2 
2 d
l

 3.657 104 N/m
The direction of the force between two wires is along the
radial line and attractive for currents traveling in the same
direction and repulsive for currents traveling in opposite
directions. The forces acting on wire M are radially away
from the other two wires. By symmetry, the horizontal
components of these forces cancel and the net force is the
sum of the vertical components.

FM  FMP cos 30  FMN cos 30

  2 3.657 104 N/m cos30  6.3104 N/m at 90

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238

Chapter 28 Sources of Magnetic Field

The force on wire N is found by adding the components of the forces from the other two wires. By
symmetry we see that this force is directed at an angle of 300. The force on wire P, will have the
same magnitude but be directed at 240.

 FN, x  FNP  FNM cos 60  3.657 104 N/m  3.657 104 N/m cos 60  1.829 104 N/m

 FN, y  FNM sin 60   3.657 104 N/m sin 60  3.167 104 N/m

   FN  2 2 3.7 104 N/m at 300
1.829 104 N/m 3.167 104 N/m 


FP  3.7 104 N/m at 240

52. The magnetic field at the midpoint between currents M and N is the

vector sum of the magnetic fields from each wire, given by Eq. 28-1.

Each field points perpendicularly to the line connecting the wire to the

midpoint.   
Bnet  BM  BN  BP

 BM
 BN  0 I  4 107 Tm A 8.00 A  9.143 105T
2 rM 2 0.0175 m

 BP
 0 I  4 107 Tm A 8.00 A  5.279 105T
2 rP 2
3 0.0175m

   Bnet  2
B  B2 2  1.849 104 T 4.571  105 T  4.93 104 T
net x net y 

net  tan1 Bnet y  tan1 4.210 105T  14
Bnet y 1.702 104 T

The net field points slightly below the horizontal direction.

53. For the wire to be suspended the net magnetic force must equal the y

gravitational force. Since the same current flows through the two lower 
FF
wires, the net magnetic force is the sum of the vertical components of
I1
the force from each wire, given by Eq. 28-2. We solve for the unknown

current by setting this force equal to the weight of the wire.

 FM 0 I M I g 2 L
 2 2 r NP l cos30  1  d l mg
4 60° I3
I2
IM   g 2rd 2 x
40I NPcos30 

       2
8900 kg m3 9.80 m s2  2 0.035m 1.00 103 m
4 4 107 Tm A 40.0 Acos30  170 A

54. The centripetal force is caused by the magnetic field, and is given by Eq. 27-5b. From this force we

can calculate the radius of curvature.

F  qvB sin  qv B  m v2 
r

     r
 mv  9.111031kg 1.3107 m s sin 7o  2.734 104 m  0.27 mm
qB 1.60 1019 C 3.3 102 T

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239

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

The component of the velocity that is parallel to the magnetic field is unchanged, and so the pitch is

that velocity component times the period of the circular motion.

2 r 2 mv 2 m
v qB qB
T   
v

      p
 vT  v cos 7  2 m   1.3107 m s cos 7 2 9.111031kg  1.4 cm
 qB  1.60 1019 C 3.3102 T

55. (a) Use Eq. 28-1 to calculate the field due to a long straight wire.

 BA at B
 0 I A  4 107 Tm A 2.0 A  2.667 106 T  2.7 106 T
2 rA to B 2 0.15m

 (b)
BB at A  0 I B  4 107 Tm A 4.0 A  5.333106 T  5.3106 T
2 rB to A 2 0.15m

(c) The two fields are not equal and opposite. Each individual field is due to a single wire, and has

no dependence on the other wire. The magnitude of current in the second wire has nothing to

do with the value of the field caused by the first wire.

(d) Use Eq. 28-2 to calculate the force due to one wire on another. The forces are attractive since

the currents are in the same direction.

     F  F  on A due to B
on B due to A 0 IAIB  4 107 Tm A 2.0 A 4.0 A
0.15 m
 lA lB 2 dA to B 2

 1.067 105 N m  1.1105 N m

These two forces per unit length are equal and opposite because they are a Newton’s third law
pair of forces.

56. (a) The magnetic field from the long straight wire will be out of the page
in the region of the wire loop with its magnitude given by Eq. 28-1.
By symmetry, the forces from the two horizontal segments are
equal and opposite, therefore they do not contribute to the net force.
We use Eq. 28-2 to find the force on the two vertical segments of
the loop and sum the results to determine the net force. Note
that the segment with the current parallel to the straight wire
will be attracted to the wire, while the segment with the
current flowing in the opposite direction will be repelled from the wire.

Fnet  F2  F4   0 I1 I 2 l  0 I1I 2 l  0 I1 I 2 l  1  1 
2 d2 2 d1 2  d1 d2 
 

 
4 107 Tm A  2.0 A 10.0 A  0.26 m  1  1 
2  0.05 0.12 
m m

 1.2 105 N toward the wire
(b) Since the forces on each segment lie in the same plane, the net torque on the loop is zero.

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240

Chapter 28 Sources of Magnetic Field

57. The sheet may be treated as an infinite number of parallel wires. D
The magnetic field at a location y above the wire will be the sum of B
the magnetic fields produced by each of the wires. If we consider
the magnetic field from two wires placed symmetrically on either y 
j
side of where we are measuring the magnetic field, we see that the t
vertical magnetic field components cancel each other out.

Therefore, the field above the wire must be horizontal and to the y

left. By symmetry, the field a location y below the wire must have 
B
the same magnitude, but point in the opposite direction. We

calculate the magnetic field using Ampere’s law with a rectangular

loop that extends adistance y above andbelow the current sheet, as shown in the figure.

 Bd l   Bd l   Bd l   Bd l  0  2BD  0Iencl  0  jtD
sides top bottom

 B  1 0 jt, to the left above the sheet
2

58. (a) We set the magnetic force, using Eq. 28-2, equal to the weight of the wire and solve for the
necessary current. The current must flow in the same direction as the upper current, for the
magnetic force to be upward.

0 I1I 2   d 2 
2 r  4 
FM  l   g   



     I2 m3 2
 g 2rd 2 8900 kg 2 9.80 m s2  2 0.050 m 1.00 103 m
 40 I1  4 107 Tm A 48.0 A  360 A, right

(b) The lower wire is in unstable equilibrium, since if it is raised slightly from equilibrium, the

magnetic force would be increased, causing the wire to move further from equilibrium.
(c) If the wire is suspended above the first wire at the same distance, the same current is needed,

but in the opposite direction, as the wire must be repelled from the lower wire to remain in

equilibrium. Therefore the current must be 360 A to the left. This is a stable equilibrium for

vertical displacement since if the wire is moved slightly off the equilibrium point the magnetic
force will increase or decrease to push the wire back to the equilibrium height.

59. The magnetic field at the center of the square loop is four times the magnetic field from one of the

sides. It will be directed out of the page. We can use the result of Problem 40 for the magnitude of

the field from one side, with R  1 d . If the current is flowing counterclockwise around the square
2

loop, the magnetic field due to each piece will point upwards.

d kˆ d kˆ
4R2  d 2
      0 I  0 I 1/2  0I kˆ
2 R 2 d
Bone 1/ 2 2 1 d
wire 2
4 1 d 2 d2
2

 2 20I kˆ
Btotal  4Bone  d

wire

60. The magnetic field at the center of a circular loop was calculated in Example 28-12. To determine
the radius of the loop, we set the circumferences of the loops equal.

2 R  4d  R  2d ; Bcircle  0 I  0 I 2 2 0 I  Bsquare
 2R 4d d

Therefore, changing the shape to a circular loop will decrease the magnetic field.

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241

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

61. (a) Choose x = 0 at the center of one coil. The center of the other coil will then be at x = R. Since

the currents flow in the same direction in both coils, the right-hand-rule shows that the magnetic
fields from the two coils will point in the same direction along the axis. The magnetic field
from a current loop was found in Example 28-12. Adding the two magnetic fields together
yields the total field.

B(x)  0 NIR 2  0 NIR 2

2 R2  x 2 3 / 2 2  R2   x  R 2  3 / 2
 


(b) Evaluate the derivative of the magnetic field at x 1 R.
2

dB   30 NIR2 x  30NIR2 ( x  R)   30 NIR3  30 NIR 3  0
dx  R 2  x2 5 R2  R2 / 45 4 R2  R2 / 45/ 2
2 / 2 2  R 2   x  R2 5/ 2 4  / 2
 

Evaluate the second derivative of the magnetic field at x 1 R.
2

d 2B   30 NIR 2  150NIR2 x2  30 NIR 2  150NIR2 ( x  R)2
dx 2
2  R 2  x2 5/ 2 2 R2  x 2 7 / 2 2  R 2   x  R2 5/2 2  R2   x  R 2 7 / 2

   
 

  30 NIR 2  150 NIR 4  30 NIR 2  150 NIR 4
2 5R2 / 45/ 2 8 5R2 / 47/ 2 2 5R2 / 45/ 2 8 5R2 / 47 / 2

 0 NIR 2   3  15 4  3  15 4   0
5R2 / 45/ 2  2 85 2 85 

Therefore, at the midpoint dB 0 and d2B  0.
dx dx2

(c) We insert the given data into the magnetic field equation to calculate the field at the midpoint.

0 NIR 2 0 NIR 2 0 NIR 2
       B 1 R   
2 2 2 3/ 2 2 2 3/ 2 R2 2 3/ 2
2  R 1 R  2  R 1 R   1 R 
  2   2   2

107 Tm A 2502.0 A0.10 m2
  4  4.5 mT

0.10 m2   0.05 m 2 3/ 2


62. The total field is the vector sum of the fields from the two currents. We

can thereBfodrel w ritBe1thdelpath integral as the sum of two such integrals.
B2 d l

To evaluate the integral for current 1, we use Eq. 28-1, with the

magnetic field constant and parallel to the loop at each line segment.

    0 I1 2 rd  0 I1
B1d l 2 r 0

To evaluate the integral for current 2, we consider a different angle d

centered at I2 and crossing the path of the loop at two locations, as
shown in the diagram. If we integrate clockwise around the path, the

components of dl parallel to the field will be r1d and r2d.
Multiplying these components by the magnetic field at both locations

gives the contribution to the integral from the sum of these segments.

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242

Chapter 28 Sources of Magnetic Field

B1d l1  B2d l2  0 I 2  r1d   0 I 2  r2 d   0
2 r1 2 r2

The totaBlidnltegralBw1 idlllbe thBe sum of these pairs resulting in a zero net integral.
2 d l  0I1  0  0I1

63. From Example 28-12, the magnetic field on the axis of a circular loop of wire of radius R carrying

B 0 IR2 , where x is the distance along the axis from the center of the loop.
2 R2  x2
 current I is 3/ 2

For the loop described in this problem, we have R  x  REarth.

R2  x2 3/ 2 2B 3/2
     
0 IR2 2B R  R2 2 2 2 BR3/ 2
R2  x2 I EarthEarth  Earth
 B  
2 3/2 0R2  R2 0
0 Earth

   223/2 1104T 6.38 106 m  3109 A

 4 107 Tm A

64. The magnetic field from the wire at the location of the plane is perpendicular to the velocity of the

plane since the plane is flying parallel to the wire. We calculate the force on the plane, and thus the

acceleration, using Eq. 27-5b, with the magnetic field of the wire given by Eq. 28-1.

F  qvB  qv 0 I
2 r

   a
 F  qv 0 I  18 103 C 2.8 m/s 4 107 Tm/A 25 A
m m 2 r 2 0.175 kg0.086 m

 1.67 105 m/s2  1.7 106 g ' s

65. (a) To find the length of wire that will give the coil sufficient resistance to run at maximum power,

we write the power equation (Eq. 25-7b) with the resistance given by Eq. 25-3. We divide the

length by the circumference of one coil to determine the number of turns.

V2 V2 V 2d2
R l d2  Pmax
 Pmax   l 

35 V2 2.0 103 m2
m 1.68 108 m 1.0 103
    N
 l V 2d2   2.0 W  46 turns
D   D Pmax

(b) We use the result of Example 28-12 to determine the magnetic field at the center of the coil,

with the current obtained from Eq. 25-7b.

   B 
0 NI  0 N Pmax  4 107 Tm/A 46 1.0 103 W  0.83 mT
D D V 2.0 m35 V

(c) Increasing the number of turns will proportionately increase the resistance and therefore
decrease the current. The net result is no change in the magnetic field.

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243

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

66. The magnetic field at the center of the square is the vector sum of the

magnetic field created by each current. Since the magnitudes of the

currents are equal and the distance from each corner to the center is the

same, the magnitude of the magnetic field from each wire is the same

and is given by Eq. 28-1. The direction of the magnetic field is

directed by the right-hand-rule and is shown in the diagram. By

symmetry, we see that the vertical components of the magnetic field

cancel and the horizontal components add.
    
B  B1  B2  B3  B4  4  0 I  cos 45ˆi
 2 r 


 
 4  0 I  2 ˆi   2 0 I ˆi
 2 d
 2 2 d
2

67. The wire can be broken down into five segments: the two a

long wires, the left vertical segment, the right vertical

segment, and the top horizontal segment. Since the current I3

in the two long wires either flow radially toward or away

from the point P, they will not contribute to the magnetic a a

field. The magnetic field from the top horizontal segment I2

points into the page and is obtained from the solution I4

to Problem 40. I1 P I5

0 I a 0 I
2 a a2  4a2 2 a 5
 Btop 1 
2

The magnetic fields from the two vertical segments both point into the page with magnitudes

obtained from the solution to Problem 41.

0 I a  0I
a 5
4 a / 2 a2  a / 22
 Bvert 1
2

Summing the magnetic fields from all the segments yields the net field.

B  Btop  2Bvert  0I  2 0I  0I 5 , into the page.
2 a 5  a 5 2 a

68. Use Eq. 28-4 for the field inside a solenoid.

 B  0IN  4 107 Tm A 2.0 A 420  8.8 103T
l 0.12 m

69. The field due to the solenoid is given by Eq. 28-4. Since the field due to the solenoid is

perpendicular to the current in the wire, Eq. 27-2 can be used to find the force on the wire segment.

          F  I l B  I l  lI N wire wire solenoid
0 solenoid 22 A 0.030m 4 107 Tm A 33A 550
solenoid 0.15 m
wire wire

 0.10 N to the south

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244

Chapter 28 Sources of Magnetic Field

70. Since the mass of copper is fixed and the density is fixed, the volume is fixed, and we designate it as

VCu  mCu Cu  lCu ACu . We call the fixed voltage V0 . The magnetic field in the solenoid is given by
Eq. 28-4.

B  0IN  0V0 N  0V0 N  0V0 N ACu  0V0 N mCu Cu
lsol RCu lsol  R Cu  lRCu sol lCu  R Cu l l2
lCu lsol
ACu sol Cu

 0V0mCu Cu N
 R Cu l l2

sol Cu

The number of turns of wire is the length of wire divided by the circumference of the solenoid.

lCu

N  lCu  B 0V0mCu Cu N  0V0mCu Cu 2 rsol  0V0mCu Cu 1
2 rsol  R Cu l l2  R Cu l l2 2RCu l r lsol sol Cu

sol Cu sol Cu

The first factor in the expression for B is made of constants, so we have B  1 . Thus we
l r lsol sol Cu

want the wire to be short and fat . Also the radius of the solenoid should be small and the length of

the solenoid small.

71. The magnetic field inside the smaller solenoid will equal the sum of the fields from both solenoids.
The field outside the inner solenoid will equal the field produced by the outer solenoid only. We set

the sum of the two fields given by Eq. 28-4 equal to  1 times the field of the outer solenoid and
2

solve for the ratio of the turn density.

0 I  na  0 Inb   1  0Inb   nb  2
2 na 3

72. Take the origin of coordinates to be at the center of the semicircle.The magnetic field at the center of

the semicircle is the vector sum of the magnetic fields from each of the two long wires and from the

semicircle. By the right-hand-rule each of these fields point into the page, so we can sum the

magnitudes of the fields. The magnetic field for each of the long segments is obtained by

integrating Eq. 28-5 over the straight segment.
 
l dlR dx ˆi  xˆi  rˆj
x2  r2 3/2
  0 I d  Rˆ  0 I R3  0 I 0   0 Ir kˆ 0 dx
4 4 4  4  x2  r2
       B straightR2 3/2

0

0 Ir kˆ x  0 I kˆ
4 x2  r2 1/2 4 r
  r 2 



magnetic field is obtained by integrating over the semicircle.
 The  0I kˆ
dfolrr2thrˆecur4ve0drI2sekˆg0mr desnt 4r Eq. 28-5

  0 I r
Bcurve 4 0

   2 0 I kˆ  0 I kˆ   0 I  2    kˆ
B  2Bstraight  Bcurve 4 r 4r 4 r

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245

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

73. (a) Set x = 0 at the midpoint on the axis between the two loops. Since the loops are a distance R

apart, the center of one loop will be at x   1 R and the center of the other at x  1 R. The
2 2

currents in the loops flow in opposite directions, so by the right-hand-rule the magnetic fields

from the two wires will subtract from each other. The magnitude of each field can be obtained

from Example 28-12.

B(x)  0 NIR 2  0 NIR 2

2  R2   1 R  x 2 3/ 2 2  R 2   1 R  x 2 3/ 2
 2   2 

Factoring out 1 R3 from each of the denominators yields the desired equation.
8

B(x)  40 NI  40 NI

R 4  1  2 x / R 2  3/ 2 R  4  1  2 x / R 2  3/ 2
  

40 NI 4 2x 2  3 / 2 4 2x 2 3/ 2 
R  R    R   
  1     1   

(b) For small values of x, we can use the approximation 1  2x 2 1 4x .
R  R

B(x)  40 NI 4 1 4x 3/ 2  4 1 4x 3/ 2 
R R  R  


 40 NI 1  4x 3/ 2  1  4x 3/ 2 
5R 5 5R  5R  


Again we can use the expansion for small deviations 1  4x 3/ 2 1 6x
5R  5R

B(x)  40 NI 1  6x   1  6x   480 NIx
5R 5 5R  5R  25R2 5

This magnetic field has the expected linear dependence on x with a coefficient of

 C  480NI / 25R2 5 .

(c) Set C equal to 0.15 T/m and solve for the current.

25CR2 5 250.15 T/m0.04 m2 5
480 N 48 4 107 T  m/A 150
 I   1.5 A

74. We calculate the peak current using Eqs. 25-7 and 25-9. Then we use the peak current in Eq. 28-1 to

calculate the maximum magnetic field.

I max  2Irms  2 Pavg  Bmax  0 Imax  20 Pavg
Vrms 2 r 2 r Vrms

  2 4 107 Tm/A  45  106 W  71 T
15  103 V
2 12 m

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246

Chapter 28 Sources of Magnetic Field

75. We use the results of Example B (10-3 T) 2.5
28-12 to calculate the magnetic field as 2.0
a function of position. The spreadsheet 1.5
used for this problem can be found on 1.0
the Media Manager, with filename 0.5
“PSE4_ISM_CH28.XLS,” on tab
“Problem 28.75.”

0.0 40
10
-40 -30 -20 -10 0 10 20 30

x (cm)

4 107 Tm/A 2.0A0.15m2
 B  250 7.0686 106 Tm3
 2
N 0IR2   0.15 m 2 x2 3/ 2   0.15 m 2 x2 3/ 2 .
R2  x2 3/2  
2  

76. (a) Use the results of Problem 61(a) to write the magnetic field.

B(x)  0 NIR 2  0 NIR 2
2 R2  x2 3/ 2
2  R 2   x  R 2 3 / 2
 

(b) See the graph. The spreadsheet 4.55
used for this problem can be

found on the Media Manager, 4.50

with filename B (10-3 T) 4.45
“PSE4_ISM_CH28.XLS,” on tab 4.40
“Problem 28.76b.” 4.35

4.30

4.25 2 46 8
0
x (cm)

(c) Use the values from the spreadsheet to find the % difference.

% diff  Bx  6.0cm  B  x  5.0cm 100  4.49537 mT  4.49588mT 100
B  x  5.0cm 4.49588 mT

 1.1102%

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247

CHAPTER 29: Electromagnetic Induction and Faraday’s Law

Responses to Questions

1. Using coils with many (N) turns increases the values of the quantities to be experimentally
measured, because the induced emf and therefore the induced current are proportional to N.

2. Magnetic flux is a quantitative measure of the number of magnetic field lines passing through a
given area. It depends not only on the field itself, but also on the area and on the angle between the
field and the area.

3. Yes, the current is induced clockwise. No, there is no induced current if the magnet is steady,
because there is no changing flux through the ring. Yes, the current is induced counterclockwise.

4. There is no induced current in the loop that is moving parallel to the wire because there is no change
of magnetic flux through the loop. The induced current in the loop moving away from the wire is
clockwise. The magnetic field through the loop due to the current is directed into the page, and the
loop is moving such that its distance from the wire is increasing, resulting in a decrease in magnetic
field strength and therefore a decrease in magnetic flux through the loop. By Lenz’s law, a
decreasing magnetic flux into the page results in a clockwise induced current.

5. Yes. The force is attractive. The induced clockwise current in the right loop will induce a
counterclockwise current in the left loop which will slow the relative motion of the loops.

6. (a) Yes.
(b) The current starts as soon as the battery is connected and current begins to flow in the first loop.
(c) The induced current stops as soon as the current in the first loop has reached its steady value.
(d) The induced current in the second loop will be counterclockwise, in order to oppose the
change.
(e) While there is an induced current, there will be a force between the two loops.
(f) The force will be repulsive, since the currents are in opposite directions.

7. Yes, a current will be induced in the second coil. It will start when the battery is disconnected from
the first coil and stop when the current falls to zero in the first coil. The current in the second loop
will be clockwise.

8. Counterclockwise. If the area of the loop decreases, the flux through the loop (directed out of the
page) decreases. By Lenz’s law, the resulting induced current will be counterclockwise to oppose the
change. Another way to approach this question is to use the right-hand rule. As the bar moves to the
left, the negative electrons in the bar will experience a force down, which results in a
counterclockwise current.

9. (a) The current through RA will be to the right. The field due to the current in coil B will be to the
left. As coil B is moved toward coil A, the flux through A will increase, so the induced field in
coil A will be to the right, to oppose the change. This field corresponds to an induced current
flowing from left to right in RA.

(b) The current through RA will be to the left. When coil B is moved away from coil A, the flux
through coil A will decrease, so the induced field will be to the left, to oppose the change. This
field corresponds to an induced current flowing from right to left in RA.

(c) If RB is increased, the current in the circuit will decrease, decreasing the flux through coil A,
resulting in a current through RA to the left.

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248

Chapter 29 Electromagnetic Induction and Faraday’s Law

10. The shielding prevents external fields from inducing a current which would cause a false signal in
the inner signal wire.

11. The currents in the two wires will be 180º out of phase. If they are very close together, or wrapped
around each other, then the magnetic fields created by the currents in the wires will very nearly
cancel each other.

12. The straight wire will fall faster. Since the magnetic field is non-uniform, the flux through the loop
will change as the loop falls, inducing a current which will oppose the change and therefore resist
the downward motion. Eddy currents will also be induced in the straight wire, but they will be much
smaller since the straight wire does not form a closed loop.

13. (a) Yes. If a rapidly changing magnetic field exists outside, then currents will be induced in the
metal sheet. These currents will create magnetic fields which will partially cancel the external
fields.

(b) Yes. Since the metal sheet is permeable, it will partially shield the interior from the exterior
static magnetic field; some of the magnetic field lines will travel through the metal sheet.

(c) The superconducting sheet will shield the interior from magnetic fields.

14. Each of the devices mentioned has a different operating current and voltage, and each needs its own
transformer with its own ratio of primary to secondary turns designed to convert normal household
current and voltage into the required current and voltage. If the devices were designed to operate
with the same current and voltage, they could all run on identical transformers.

15. You could hook the transformer up to a known ac voltage source. The ratio of the output voltage to
the input voltage will give the ratio of turns on the two coils. If you pair up the leads incorrectly (one
lead from each coil, rather than both leads from the same coil), there will be no output voltage.
Alternatively, you could attach an ohmmeter to two of the leads. The resistance will be infinite if
you have one lead from each pair, and nearly zero if you have both leads from the same pair.

16. Higher voltages are inherently more dangerous because of the increased risk of establishing large
currents and large electromagnetic fields. The large potential differences between the wires and the
ground could cause arcing and short circuits, leading to accidental electrocutions. In addition,
higher-voltage power lines will have higher electromagnetic fields associated with them than lower-
voltage power lines. Biological effects of exposure to high electromagnetic fields are not well
understood, but there is evidence of increased health risks to people who live close to high voltage
power lines.

17. When the transformer is connected to the 120-V dc source no back emf is generated, as would
happen with an ac source. Therefore, the current in the transformer connected to the dc source will
be very large. Because transformers generally are made with fine, low resistance wires, the large
current could cause the wires to overheat, melt the insulation, and burn out.

18. A motor uses electric energy to create mechanical energy. When a large electric motor is running,
the current in the motor’s coil creates a back emf. When the motor is first turned on, the back emf is
small, allowing the motor to draw maximum current. The back emf has a maximum value when the
motor is running at full speed, reducing the amount of current required to run the motor. As the
current flow in the motor’s coil stabilizes, the motor will operate at its lower, normal current. The
lights will dim briefly when the refrigerator motor starts due to the increased current load on the
house circuit. Electric heaters operate by sending a large current through a large resistance,
generating heat. When an electric heater is turned on, the current will increase quickly to its

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249

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

maximum value (no coil, so no back emf) and will stay at its maximum value as long as the heater is
on. Therefore, the lights will stay dim as long as the heater is on.

19. At the moment shown in Figure 29-15, the armature is rotating clockwise and so the current in
length b of the wire loop on the armature is directed outward. (Use the right-hand rule: the field is
north to south and the wire is moving with a component downward, therefore force on positive
charge carriers is out.) This current is increasing, because as the wire moves down, the downward
component of the velocity increases. As the current increases, the flux through the loop also
increases, and therefore there is an induced emf to oppose this change. The induced emf opposes the
current flowing in section b of the wire, and therefore creates a counter-torque.

20. Eddy currents exist in any conducting material, so eddy current brakes could work with wheels made
of copper or aluminum.

21. The nonferrous materials are not magnetic but they are conducting. As they pass by the permanent
magnets, eddy currents will be induced in them. The eddy currents provide a “braking” mechanism
which will cause the metallic materials to slide more slowly down the incline than the nonmetallic
materials. The nonmetallic materials will reach the bottom with larger speeds. The nonmetallic
materials can therefore be separated from the metallic, nonferrous materials by placing bins at
different distances from the bottom of the incline. The closest bin will catch the metallic materials,
since their projectile velocities off the end of the incline will be small. The bin for the nonmetallic
materials should be placed farther away to catch the higher-velocity projectiles.

22. The slots in the metal bar prevent the formation of large eddy currents, which would slow the bar’s
fall through the region of magnetic field.

23. As the aluminum sheet is moved through the magnetic field, eddy currents are created in the sheet.
The magnetic force on these induced currents opposes the motion. Thus it requires some force to
pull the sheet out. (See Figure 29-21.)

24. As the bar magnet falls, it sets up eddy currents in the metal tube which will interact with the magnet
and slow its fall. The magnet will reach terminal velocity (due to the interactions with the magnetic
dipoles set up by the eddy currents, not air resistance) when the weight of the magnet is balanced by
the upward force from the eddy currents.

25. As the bar moves in the magnetic field, induced eddy currents are created in the bar. The magnetic
field exerts a force on these currents that opposes the motion of the bar. (See Figure 29-21.)

26. Although in principle you could use a loudspeaker in reverse as a microphone, it would probably not
work in actual practice. The membrane of the microphone is very lightweight and sensitive to the
sound waves produced by your voice. The cardboard cone of a loudspeaker is much stiffer and
would significantly dampen the vibrations so that the frequency of the impinging sound waves
would not be translated into an induced emf with the same frequency.

Solutions to Problems

1. The average induced emf is given by Eq. 29-2b.

e  N dB  N B  2 38 Wb  58 Wb  460 V
dt t 0.42 s

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