Module NF2FSLT SP025

Solution IX  50 A;d  5.0102 m; IY 10 A

b. μ0 I X μ0 I Y
2πrX 2πrY


IX  IY
r
d  r

 50 

r
10
5.0 102  r

r  4.1667 102 m

LO 4.2 Resultant magnetic field produced by current-carrying conductor

Exercise 56

2 long parallel wires carry currents of
and 2 A. What is the resultant magnetic
field at:
(i) midway between the wires ?
(ii) 1.5 cm to the left of the 1st wire ?

Solution At midway, the resultant magnetic
field is due to magnetic field

produced by wire 1 and wire 2.

(i) midway between the wri1rers2  12  6 cm
2

For 1st wire :

B1  oI 1  4 107 (8)  2.6667 105 T
2 r1 2 (6 102 )

For 2nd wire : into the page ( x )

B2  oI 2  4 107 (2)  6.6667 106 T
2 r 2 2 (6 102 )

out of page ( )

Solution

The resultant magnetic field at that point :

Minus because B1 and B2 in opposite direction

 

 B  B1  B2



  B  2.6667105   6.6667106

 B  2105 T

Direction : into the page ( X )

For 1st wire :

(ii) 1.5 cm to the left of 1st wire :  oI 1  4 107 (8)  1.0667 104
2 r1 2 (1.5102 )
B1 = ( • ) B1 T
B2 = ( • )
out of page ( )

For 2nd wire :

B2  oI 2  4 107 (2)  2.963106 T
2 r2 2 (13.5102 )
out of page ( )

 

The resultant magnetic field atthat point : B  B1  B2
 B  1.0667 104  (2.963106 )

 1.0963104 T

Direction : out of page ( )

Where on a line perpendicular to The point must
lie between 1st
and joining the wires is the wire & 2nd wire
where the B
magnetic field zero ? produced by
both wire are
Assume the point is at a distance x from  in opposite
1st wire: direction and
B  0 will cancel out
B1  B2 each others.

I1= 8 A I2 = 2 A oI1  oI 2
2 r1 2 r2
12 cm

x 0.12–x 8 2
x (0.12  x)
B1()  0.12  x  0.25x
B2()

x  0.12  0.096 m
1.25

LO 4.2 Resultant magnetic field produced by current-carrying conductor

Exercise 57

Two long straight wires are oriented
perpendicular to the page as shown in
figure (a).
The current in one wire is I1 = 3.0 A
pointing into the page and the current in
the other wire is I2= 4.0 A pointing out of
page.

Determine the magnitude and direction of
the net magnetic field intensity at point P.

Solution I1  3.0 A ; I2  4.0 A ; r1  5.0102 m

P     r2  5.0 102 2  5.0 102 2
θ
B1

 r1 θ r2 r2  7.0711102 m
B2 I1 X 5.0 102 m
5.0 10 2
I2 tan θ  5.0 10 2

θ  45

By applying the equation of magnetic field intensity for straight wire, thus

μ0 I1    B1
B1  2πr1  4π 107 3.0 B1  1.20 105 T
2π 5.0 102
B2  1.1314105 T
μ0 I 2    B2
B2  2πr2  4π 107 4.0
2π 7.1102

Vector x-component (T) y-component (T)

 B1  1.20 105 T 0
B1
 B2 sin θ
 B2 cosθ
  1.1314105 cos 45    1.1314105 sin 45

 8.002106
 B2
By  0  8.0002 106
 8.002106
 8.002 106
Bx  1.20105  8.0002106

 3.9998106

Therefore the magnitude of the net magnetic field intensity at point P is given by

   B  Bx2  By2  3.9998106 2   8.0002106 2 B  8.9444 106 T

and its direction is θ  tan 1 By   
Bx B2 P B1

 tan 1  8.0002 106  θ  63.44 below positive x - axis 63.1
3.9998 106 B

LO 4.2 Resultant magnetic field produced by current-carrying conductor

Exercise 58

A circular coil has 15 turns and a diameter of 45.0 cm. If the magnetic
field strength at the center of the coil is 8.0×10−4 T, find the current

flowing in the coil.

Solution B  oNI

Given: 2r
Magnetic field, B= 8.0 x 10-4 T
Number of turns, N = 15 turns     I
Diameter, D = 45.0 cm = 45.0 x 10-2 m  2 8104 22.5102
Radius, R = 22.5 cm =22.5 x 10-2 m
4 107 15

I  19.0986 A

LO 4.2 Resultant magnetic field produced by current-carrying conductor

Exercise 59

A long straight wire carries a current of I1 = 8.0A lies next to a 
BC  ?
circular loop of radius r = 0.03 m that carries a current I2 = 2.0

A. Find the magnitude and direction of the magnetic field at the

center C of the loop.

Solution The net magnetic field Apply RHR-2 shows that at point C
at the point C is the
sum of two the field B1 is directed upward.
contributions: (1) the The magnetic field B2 is directed
field B1 produced by downward, opposite to the direction
the long, straight wire,
and (2) the field B2 of B1.
produced by the
circular loop.

Solution Magnitude and direction of the magnetic field

Long, at the center C of the loop, Bc = ?
straight
  Minus because
wire BC  B1  B2

B1 and B2 have
opposite

oI 1    oI 2  direction
2rlsw
 2rc Center of a
circular loop

 4 107 (8)  4 107 (2)
 2 (0.03) 2(0.03)

BC  1.1445 T

The net field is +, so it is directed upward

LO 4.2 Resultant magnetic field produced by current-carrying conductor

Exercise 60

A closely wound circular coil of diameter 10 cm has 500 turns and carries a

current of 2.5 A. Determine the magnitude of the magnetic field at the

centre of the coil. (Given 0 = 4  107 T m A1)

Solution r  10102  5102 m; N  500; I  2.5 A
2

By applying the equation for magnitude of the magnetic field at the centre of the

circular coil, thus    B 
4π 107 5002.5
B  μ0 NI
2r 2 5.0 102

B  1.5708 102 T

LO 4.2 Resultant magnetic field produced by current-carrying conductor

Exercise 61

A solenoid of 50 turns is carrying a current of 10 mA. The magnetic field
strength at the centre is 1.05×10−6 T. Calculate the length of the solenoid.

Solution Given:

Magnetic field at the centre , Bcentre = 1.05 x 10-6 T
Number of turns, N = 50 turns

Current, I = 10 mA= 10 x 10-3 m

  o NI
Bcentre
L
   L 
4 107 50 10 103

1.05 106

L  0.5984 m

LO 4.2 Resultant magnetic field produced by current-carrying conductor

Exercise 62

A solenoid of length 1.5 m and 2.6 cm in diameter carries a current of
18 A. The magnetic field inside the solenoid is 2.3 mT. Calculate the
length of the wire forming the solenoid.

(Given 0 = 4  107 T m A1)

Solution Given L  1.5 m; r  2.6 102  1.3102 m; Bi  2.3103 T;
I  18 A 2

By applying the equation of magnetic flux density inside the

solenoid, thus

Bi  μ0 NI  2.3103  4π 107 N 18
l 1.5

N  153 turns

Solution

Since the shaped for each coil in the solenoid is circle, then the circumference for

one turn is

circumference  2πr  circumference  2π 1.3102

circumference  8.1681102 m

Therefore the length of the wire forming the solenoid is

x  N circumference

 x  153 8.1681102

x  12.4972 m

LO 4.2 Resultant magnetic field produced by current-carrying conductor

Exercise 63

Determine the direction of the magnetic force on the particles as they
enter the magnetic field as shown in Figure below.

Solution Applying RHR1 :

The force is
upward

Solution

Out of the plane of the
paper

No deflection

LO 4.2 Resultant magnetic field produced by current-carrying conductor

Exercise 64

Determine the direction of the magnetic force, F exerted on a charge in the following

problems:

a.  b.  
v B v

 

B d. v

c.

X X XX I

 e. 

X X X X v
v
BX X X X I

Solution 
F
a. Applying RHR1,

b. Applying RHR1,  v (into the page)
c. Applying RHR1,

B
  
(to the left) F B v



X X X X
(to the left) F 

X X X X
v
BX X X X

Solution

d. Apply RHR2 to determine the direction of magnetic

field produces by the current I on the charge position. Then

apply the RHR1, thus 
(to the left)  X XX XB
F
X I v X X

e. Apply RHR2 to determine the direction of magnetic field forms by

the current I on the charge position. Then apply the RHR1 , thus
F (upwards)
X XXBv

X



I XX

XX

LO 4.2 Resultant magnetic field produced by current-carrying conductor

Exercise 65

A beam of protons ( q = 1.6×10–19 C ) moves at 3×105 m s–1 through a uniform magnetic
field with magnitude 2.0 T that is directed along the positive z axis. The velocity of each
proton lies in the x-z plane at an angle 30° to the +z axis.
Find the force on a proton.

+ x

30º

   3 105 m s 1
v

z B2T

Solution

Given: v = 3 X 105 ms-1; B = 2.0T; θ = 30°

F  qv  B
F  qvB sin

   1.61019 3105 2.0sin 30

 4.81014 N

Using Right Hand Rule :

+ x From Right Hand Rule,
force on the proton is
30º

directed downwards @ -y

   3 105 m s 1 axis.
v

z B2T

LO 4.2 Resultant magnetic field produced by current-carrying conductor

Exercise 66

Calculate the magnitude of the force on a proton travelling 5.0107
m s1 in the uniform magnetic flux density of 1.5 Wb m2, if :
a. the velocity of the proton is perpendicular to the

magnetic field.
b. the velocity of the proton makes an angle 50 with

the magnetic field.
(Given the charge of the proton is +1.601019 C)

Solution

a. Given v  5.0107 m s1; B  1.5 Wb m2 ,   90
Therefore
F  qvBsin θ

   1.601019 5.0107 1.5sin 90

F  1.20 1011 N

b. Given   50
Hence
  F  1.601019 5.0107 1.5sin 50

F  9.19251012 N

LO 4.3 Force on a moving charged particle in a uniform magnetic field

Exercise 67

A particle with a charge of −5×10−4 C and a mass of 2×10−9kg
moves at a speed of 1×103 m s−1 in the +x direction toward a
uniform magnetic field of 0.20 T as shown in figure below.

Solution

(a) Sketch the possible path of the electron after it

enters the magnetic field ?


v

 
F 
F F v Electron
travels anti-
F clockwise
in circular
 trajectory
v

Solution (c) What is the radius of the
circular orbit of the particle
(b) What is the force acting on while it is in the field ?
the particle as it enters the
magnetic field ? Since electron travel in
circular path – the magnetic
Using: force supplies the centripetal

F  qvB sin  force. FB  Fc

 (5104 )(1103)(0.2)sin 90 F  mv2
r
F  0.10N
0.10  2 109 (1103 )2
r

r  2 102 m

LO 4.3 Force on a moving charged particle in a uniform magnetic field

Exercise 68

v

A 20.0 cm B

An electron at point A in Figure 6.26 has a speed v of 2.50
106 m s-1. Determine

a. the magnitude and direction of the magnetic field that will

cause the electron to follow the semicircular path from A to B.

b. the time required for the electron to move from A to B.

(Given e=1.601019 C and me= 9.111031 kg)

Solution

v  2.50 106 m s 1; d  20.0 102 m

a. Since the path makes by the electron is a semicircular thus the
the magnitude of the magnetic field is given by

r  mv and r  d
Be 2
  20.0102
 2
d  mv  9.111031 2.50 106 B  1.4234104 T
2 Be B 1.60 1019

Direction of magnetic field : into the page

OR  B
B
v

A F

Solution

v  2.50 106 m s 1; d  20.0 102 m

b. The period of the electron is

v  rω and ω  2π
T
v   d  2π 
 2  T   2.50 106  20.0 102 π
T

T  2.5133107 s

Since the path is the semicircular then the time required for the

electron moves from A to B is given by

t  1T
2

 t  1 2.5133107  1.2567 107 s
2

LO 4.4 Force on a current carrying conductor in a uniform magnetic field

Exercise 69

Rank the orientation according to the magnitude of
the magnetic force exerted on the wire largest to
smallest. Determine the direction of magnetic force.

Solution

Out of page No Force To the right
(a) I perpendicular to B, F is max.
(b) I parallel to B, F = 0 N.
(c) I perpendicular to B, F is max
Ranking  a, c, b

LO 4.4 Force on a current carrying conductor in a uniform magnetic field

Exercise 70

A straight horizontal copper rod carries a current of 50.0 A
from west to east in a region between the poles of a large
electromagnet. In this region there is a horizontal magnetic
field towards the northeast with magnitude 1.20 T, as shown
in figure.Find the magnitude and direction of the force on a
1.0 m section of rod.

North


B

I  50.0 A 45  East
L
West
1.00 m
B  1.20 T

South

Solution

Magnitude of magnetic force :

F  ILB sin  (50)(1.0)(1.20) sin 450
 42.4264 N

Direction of magnetic force :

Apply RHR1 rule
 the force is out of the
plane

LO 4.4 Force on a current carrying conductor in a uniform magnetic field

Exercise 71

A 45 m length of wire is stretched horizontally between two vertical
posts. The wire carries a current of 75 A and experiences a magnetic
force of 0.15 N. Find the magnitude of the earth’s magnetic field at
the location of the wire, assuming the field makes an angle of 60° with
respect to the wire.

Solution Using: F  ILB sin

B  IL F  0.15
75(45) sin 600
sin

B  5.1320105 T

LO 4.4 Force on a current carrying conductor in a uniform magnetic field

Exercise 72
Two conducting rails are 1.6 m apart and are parallel to the ground at the same
height.

A 0.2 kg aluminum rod is lying on top of the rails and a 0.05 T
magnetic field points upward, perpendicular to the ground. There is a
current I in the rod, directed as in drawing. The coefficient of static
friction between the rod and each rail is μs = 0.45.
How much current is needed to make the rod begin moving and in

Solution For the rod begin to move, the magnetic force is
equals to the frictional force.

F  2 fs(max)

ILB sin  2 s N * N  mg

Since F pull rod to left, I  2 s (mg )  2(0.45)(0.2)(9.81)
BL sin 0.05(1.6) sin 900
friction act in opposite
direction to the right. I  22.0725 A

Rod carry current put in Apply RHR1 rule  the rod moves to left.
magnetic field, have

magnetic force, F.
Apply RHR, F is to the

left.

LO 4.5 Forces between two parallel current-carrying conductors

Exercise 73

2 long, straight, parallel wires carry current in the same direction.

(a) Determine whether the forces on the wire are attractive or
repulsive ?

(b) If the wires are 24 cm apart & carry currents of 2 A & 4 A
respectively, find the force per unit length on each wire.

Solution

(a) currents flow in same direction
– attractive forces

(b) The force per unit length on each wire :

F   I Io 1 2
L 2 r

 4 107 (2)(4)
2 (24 102 )
F  6.6667 106 N m-1
L

LO 4.5 Forces between two parallel current-carrying conductors

Exercise 74

Two long straight parallel wires are placed 0.25 m apart in a vacuum. Each wire
carries a current of 2.4 A in the same direction.
a) Sketch a labeled diagram to show clearly the direction of the force on each

wire.
b) Calculate the force per unit length between the wires.
c) If the current in one of the wires is reduced to 0.64 A, calculate the current

needed in the second wire to maintain the same force per unit length between
the wires as in (b).

Solution I1  I2  2.4 A ; d  0.25 m

a. The diagram is b. The force per unit length between the wires is given by

I1   I2 F  μ0 I1I2 F  4π 107 2.42.4
L 2πd L
B F21 F12 B 2π0.25

1 d 21 F  4.608106 N m1
L
2

c. Given I1  0.64 A

Therefore the current needed in the second wire is

 4.608106 
F  μ0 I1I2 4π 107 0.64I2
L 2πd 2π0.25

I2  9.0 A

LO 4.6 Torque on a coil

Exercise 75

A rectangle coil of dimensions 5.4 cm x 8.5 cm consists of 25 turns of
wire and carries a current of 15 mA. A 0.35 T magnetic field is applied
parallel to the plane of the coil. Calculate the magnitude of the torque on
the coil when the field makes an angle of 30° with the plane of coil.

Solution

A = 0.054 x 0.085 = 4.59 x 10-3 m2
I = 15 x 10 -3 A
B = 0.35 T
N = 25 turns
Angle between plane and B = 30o, thus θ = 60o

Using:

  N I A B sin 

 25(15103)(4.59 103)(0.35) sin 60o

 5.2173104 N m

LO 4.6 Torque on a coil

Exercise 76

A 50 turns rectangular coil with sides 10 cm  20 cm is placed
vertically in a uniform horizontal magnetic field of magnitude 2.5
T. If the current flows in the coil is 7.3 A, determine the torque
acting on the coil when the plane of the coil is
a) perpendicular to the field,
b) parallel to the field, and
c) at an angle of 75 to the field.

Solution

a) Plane of the coil perpendicular to the field,
Angle between normal and B, θ = 0o

  N I A B sin  50(7.3)(10 102 )(20 102 )(2.5) sin 00

0Nm

b) Plane of the coil parallel to the field,
Angle between normal and B, θ = 90o

  N I A B sin

 50(7.3)(10 102 )(20 102 )(2.5) sin 900

18.25 N m

Solution

c) Plane of the coil at 750 to the field,
Angle between normal and B,

  900  750  150

  N I A B sin

 50(7.3)(10 102 )(20 102 )(2.5) sin150
 4.7234 N m

LO 4.7 Application of Motion of charger particle

Exercise 77

An electron with kinetic energy of 8.01016 J passes perpendicular
through a uniform magnetic field of 0.40103 T. It is found to follow a
circular path. Calculate

a. The speed of the electron.
b. the radius of the circular path.
c. the time required for the electron to

complete one revolution.

(Given e/m = 1.761011 C kg-1, me = 9.111031 kg)

Solution

b. Since the path made by the
Given: K  8.0 1016 J; B  0.40 103 T electron is circular,

a. The speed of the electron is given thus F9B0FmC v2
by r
evB sin
K  1 mv2
2  e B  v
m r
 8.01016  1 9.111031 v2
2

v  4.1908107 m s1   1.761011 0.40103  4.1908107
r

r  0.5953 m

Solution

c. The time required for the electron to complete
one revolution is given by

v  2πr
T

4.1908107  2π0.5953

T
T  8.9252108 s