Module NF2FSLT SP025

LO 4.7 Application of Motion of charger particle

Exercise 78

A singly charged ion having a particular velocity is selected
using a magnetic field of 0.10 T perpendicular to an electric
field of 1×103 V m-1. This same magnetic field is used to
deflect the ion which moves in a circular path with a radius of
1.2 cm. What is the mass of the ion ?

Solution

From: q E
m rB2
m  qrB 2

E
 1.6 1019 (1.2 102 )(0.1) 2

1103

m  1.921026 kg

LO 4.7 Application of Motion of charger particle

Exercise 79

In a mass spectrometer, the magnetic fields and electric fields for
selecting the velocity of the ion are 0.56 T and 1.20 x 105 V m-1
respectively. The diameters of the circular paths for the oxygen ions
are 14.12 cm, 15.04 cm and 15.90 cm respectively. The charge of
each oxygen ions is 1.60 x 10-19 C. Calculate the mass of each type
of oxygen atoms?

Solution

In velocity selector, In evacuated chamber,

Fe  FB FB  Fc
qE  qvB
qvB  mv2
v E r
B
m  Bqr (1)
 1.2105 v
0.56

 2.1429105 m s-1

Solution

Substitute in (1) for d1 = 14.12 cm, d2 = 15.04

cm and d3 = 15.90 cm

d1 = 14.12 cm

m1  Bqr1

v 1019 )(14.12 102 )
(0.56)(1.6
2
m1  2.1429 105

 2.95201026 kg

Solution

d2 = 15.04 cm

m2  Bqr2

v )(15.04 10 2

(0.56)(1.6 1019 )
2
m2  2.1429 105

 3.14431026 kg

Solution

d3 = 15.90 cm

m3  Bqr3

v )(15.90 10 2

(0.56)(1.6 1019 )
2
m3  2.1429 105

 3.32411026 kg

LO 4.7 Application of Motion of charger particle

Exercise 80

A 0.13 MV is applied across the plates of a mass
spectrometer to determine the velocity of electrons which are
detected with photographic plate.

(a) Find the velocity of the electrons that
come out of the plates.

(b) The electrons are then entering evacuated
chamber under a centripetal force in a
circular path of radius 1.50 m. What is the
magnetic field used for this circular motion?

Solution

(a) Kinetic Energy,
1 mv2  eV
2

v  2eV
m

 (2)(1.61019 )(0.13106 )
9.111031

 2.1369108 m s-1

Solution

(b) FB  Fc
Bqv  mv2
r

B  mv
qr

 9.111031(2.1369108 )
1.6 1019 (1.50)

 8.1113104 m s-1

CHAPTER 5.0
ELECTROMAGNETIC

INDUCTION

MODE Face to face Non Face to face
SLT SLT
Lecture
Tutorial 2 2

6 6

212

CHAPTER 5
5.1 Magnetic Flux

LO 5.1 Exercise 81 Non Face-to-face SLT

A flat surface with area 3.0 cm2 is placed in a uniform
magnetic field. If the magnetic field strength is 6.0 T, making
an angle 30° with the surface area, find the magnetic flux
through this area.

2
1
3

Solution Non Face-to-face SLT

Given: B = 6.0 T, A = 3.0 cm2 = 3.0 × 10–4 m2

θ = 90 – 30 = 60°

coil

Using:

2
1
4

LO 5.1 Exercise 82 Non Face-to-face SLT

A long, straight wire carrying a current of 2.0 A is placed along
the axis of a cylinder of radius 0.50 m and a length of 3.0 m.
Determine the total magnetic flux through the cylinder.

SOLUTION:

Given: I = 2.0 A area
r = 0.5 m I
l = 3.0 m
=?

21
5

LO 5.1 Exercise 84 Non Face-to-face SLT

A solenoid 4.00 cm in diameter and 20.0 cm long has 250
turns and carries a current of 15.0 A. Calculate the magnetic
flux through the circular cross sectional area of the solenoid.

SOLUTION :

Given: d = 4.0 cm, r = 2.0cm, l = 20 cm, N= 250 turns, I = 15.0 A

NS

II

2.96×10–5 T m2 21
6

CHAPTER 5 Non Face-to-face SLT
5.2 Induced emf
Induced current
LO 5.2 flows in
counterclockwise
When magnet’s N-
pole is moving into 217
coil,

Lenz’s law

induced I flows in N
I
such a direction as to
produce a N-pole to
oppose the
approaching of
magnet N pole.

LO 5.2 Exercise 84 Non Face-to-face SLT

Indicate the direction of the induced I. Explain.

SN

I (i) When magnet’s N-pole is moving
into coil,
Lenz’s
law induced I flows in such a direction as
to produce a N-pole
to oppose the approaching of magnet.

218

I Non Face-to-face SLT

LO 5.2 Exercise 85
Indicate the direction of the induced I. Explain.

NS

(ii)

When magnet’s S-pole is leaving the coil, 219

induced I flows in such a direction as to
produce a N-pole to oppose the leaving of
magnet.

Non Face-to-face SLT

LO 5.2 Exercise 86

The solenoid in figure is moved at constant velocity
towards a fixed bar magnet. Using Lenz’s law, determine
the direction of the induced current through the resistor.

Movement of solenoid

ab

220

Solution North pole

South pole

Movement of
solenoid

N

ab

1) When solenoid bring towards magnet bar, it experience an

increasing in flux.

2) Flux change → current induced.
3) Lenz’s law, direction of current induced opposes the change.

The right end of solenoid must form north pole to oppose
incoming north pole of the bar magnet.

Non Face-to-face SLT

LO 5.2 Exercise 87

Determine the direction of the induced current along

PQ? • • • Q• • • • •

• •••• •••

• • • • GX • •

• • •• • ••

• •••• •••

• • • P• • • • •

222

LO 5.2 Non Face-to-face SLT

Consider the
arrangement shown
in figure.

Assume that R = 6 Ω, L = 1.2m & a uniform 2.50 T magnetic
field is directed into the page.
(a) At what speed should the bar be moved to produced a

current of 0.5A in the resistor.
(b) what is the direction of the induced current ?

223

Non Face-to-face SLT

Solution
(a) Using :

(b) Applying Right Hand Rule,

the direction of the induced current is from
b → a → d → c → b ( anticlockwise )

224

Non Face-to-face SLT

LO 5.2 Exercise 88

× ××××

The flexible loop has a

radius of 12 cm and is in a × × ×A × ×

magnetic field of strength
0.15 T. The loop is grasped × × × × ×

at point A and B and × ××××
stretched until its area is

nearly zero. If it takes 0.20 s × × × × ×

to close the loop, find the

magnitude of the average × × × × ×
B
induced emf in it during this
× ××××
time.

× ××××

225

Solution Non Face-to-face SLT
Apply Faraday’s law:
Final
× ××××
× × ×A × ×
× ××××
× ××××
× ××××
× ××××
× × ×B × ×
× ××××

226

Non Face-to-face SLT

LO 5.2 Exercise 89

A circular coil has 200 turns and diameter 36 cm. the
resistance of the coil is 2.0 Ω. A uniform magnetic field is
applied perpendicularly to the plane of the coil. If the field
changes uniformly from 0.5 T to 0 T in 0.8s.

(a) Find the induced emf & current in the coil while the field

is changed.
(b) Determine the direction of the current induced.

Solution

Given : = 200 , = 0.36 , = 2.0 Ω, = 0.8 , =
− = 0 − 0.5 = −0.5

227

Solution Non Face-to-face SLT

Given : = 200 , = 0.36 , = 2.0 Ω, = 0.8 , =
− = 0 − 0.5 = −0.5

The area of the coil, A =

From Faraday’s Law : ( * N & A constant, θ = 0° )

228

Non Face-to-face SLT

Using:

(b) Apply Lens law, B from induced current will try
to prevent the decrease in flux.

Thus induced current is in clockwise direction.

229

LO 5.2 Exercise 90 Non Face-to-face SLT

A circular coil of radius 20 cm is placed in an external magnetic
field of strength 0.2 T so that the plane of the coil is perpendicular
to the field. The coil is pulled out of the field in 0.30 s. Find the
average induced emf during this interval.

Solution

Given: r = 2.0 × 10-2 m, Bi = 0.2 T → Bf = 0T in t = 0.30 s

= 84 mV

Non Face-to-face SLT

LO 5.2 Exercise 91

A 25 turn circular coil of wire has a diameter of 1.0 m. It is
Placed with its axis along the direction of earth’s magnetic field
of 50.0 μT, and then in 0.20 s it is flipped 180°. An average emf

of what magnitude is generated in the coil?

Solution

Given: N = 25 turns, d = 1.0 m, B = 25 × 10-6 T, Flipped 180°
in 0.20 s

LO 5.2 Exercise 92 Non Face-to-face SLT

The plane of a rectangular coil, 5.0 cm by 8.0 cm, is

perpendicular to the direction of a magnetic field B. If the coil
has 75 turns and a total resistance of 8.0 Ω, at what rate must
The magnitude of B change to induce a current of 0.10 A in the

winding of the coil?

Solution
Given:


= − = −

= − 0


0.8
= = 75 40 × 10−4
= 2.67 −1
232

Non Face-to-face SLT

LO 5.2 Exercise 93

A narrow coil of 10 turns and diameter of 4.0 cm is placed
perpendicular to a uniform magnetic field of 1.20 T. After
0.25 s, the diameter of the coil is increased to 5.3 cm.

a. Calculate the change in the area of the coil.
b. If the coil has a resistance of 2.4 Ω, determine the

induced current in the coil.

Solution

Initial Final 233

Solution Non Face-to-face SLT

a. The change in the area of the coil is given by

234

Solution Non Face-to-face SLT

b. Given

The induced emf in the coil is

Therefore the induced current in the coil is given by

235

LO 5.2 Exercise 94 Non Face-to-face SLT

A loop of area 0.10 m2 is rotating at 60 rev/s with its axis of
rotation perpendicular to a 0.20 T magnetic field.
(a) If there are 1000 turns on the loop, what is the maximum

voltage induced in the loop?
(b) When the maximum induced voltage occurs, what is the

orientation of the loop with respect to the magnetic field?

236

Solution Non Face-to-face SLT
Given: A = 0.1 m2 ; B = 0.2 T ;
N = 1000
ω = 60 rev/s

(a) Maximum voltage induced → sin ωt = 1

(b) When the plane of the loop is parallel to the
magnetic field.

237

LO 5.2 Exercise 95 Non Face-to-face SLT

An AC generator consists of 8 turns of wire, each of
area A = 0.09 m2 & the total resistance of the wire is
12 Ω. The loop rotates in a 0.5 T magnetic field at a
constant frequency of 60 Hz.
(a) Find the maximum induced emf.
(b) What is the maximum induced current ?
(c) Determine how the induced emf & induced current vary

with time.

Solution
Given:

238

Solution Non Face-to-face SLT
Given:

c) i) The magnitude of induced emf and as well as
induced current is depends on the angle
between the field and the coil.

ii) The induced emf is an alternating voltage
because has positive value as well as
negative value.

iii) The emf induced in a coil varies sinusoidally
with time.

239

Non Face-to-face SLT

CHAPTER 5.3 SELF-INDUCTANCE
LO 5.3 Exercise 96

Induced emf of 5.0 V is developed across a coil when the
current flowing through it changes at 25 A s–1. Determine the
self inductance of the coil.
Solution

Using:
Minus sign just indicates the direction of emf ( Lenz’s Law ).
Therefore:

240

LO 5.3 Exercise 97 Non Face-to-face SLT

Calculate the magnetic flux through the area enclosed by a
300 turns, 7.20 mH coil when the current in the coil is 10 mA.

Solution
Given : N = 300 ; L = 7.20×10–3 H ; I = 10×10–3 A

Using:

241

LO 5.3 Exercise 98 Non Face-to-face SLT

(a) Calculate the self inductance of a solenoid containing 300
turns if the length of the solenoid is 25.0 cm & its cross
sectional area is 4 cm2.

(b) Calculate the self induced emf in the solenoid if the current
through it is decreasing at the rate of 50.0 A s–1.

Solution

Given : N = 300 ; l = 25×10–2 m ; A = 4×10–4 m2 ; dI/dt = - 50 A s–1

(a) Using:

242

Non Face-to-face SLT

Solution

Given : N = 300 ; l = 25×10–2 m ; A = 4×10–4 m2 ; dI/dt = - 50 A s–1

(b) Using :

243

LO 5.3 Exercise 99 Non Face-to-face SLT

The coil in an electromagnet has an inductance of 1.7 mH

and carries a constant direct current of 5.6 A. A switch is

suddenly opened, allowing the current to drops to zero over a
small interval of time, ∆t. If the magnitude of the emf induced
during this time is 7.3 V, what is ∆t ?

SOLUTION:

Given:


= −

1.7 × 10−3 0 − 5.6 = 1.30 × 10−3
= − = −
7.3

244

LO 5.3 Exercise 100 Non Face-to-face SLT

A 500 turns solenoid is 8.0 cm long. When the current in this
solenoid is increased from 0 to 0.25 A in 0.35 s the magnitude
of the induced emf is 0.012 V. Find
(a) the inductance and
(b) the cross sectional area of the solenoid.

SOLUTION:

Given:

(a) (b)

245

Non Face-to-face SLT

CHAPTER 5.4 ENERGY STORED IN INDUCTOR
LO 5.4 Exercise 102

A 400 turns solenoid has a cross sectional area 1.81×10-3 m2
and length 20 cm carrying a current of 3.4 A.
(i) Calculate the inductance of the solenoid.
(ii) Calculate the energy stored in the solenoid.
(iii) Calculate the induced emf in the solenoid if the current

drops uniformly to zero in 55 ms.
Solution

(i) Inductance of a solenoid,

247

(ii) Energy stored, Non Face-to-face SLT
(iii) induced e.m.f,
248

Non Face-to-face SLT

LO 5.4 Exercise 103

Energy stored in an inductor is 30 mJ when current
flowing through the inductor is 50 mA.
(a) Calculate the self inductance of the inductor.
(b) Calculate the new current if the energy stored

increases to 4 times the original energy.

SOLUTION:

(a) (b)

249

LO 5.4 Exercise 104 Non Face-to-face SLT

A solenoid of length 25 cm with an air-core consists of 100
turns and diameter of 2.7 cm. Calculate
a. the self-inductance of the solenoid, and

b. the energy stored in the solenoid, if the current

flows in it is 1.6 A.
(Given μo = 4π × 10−7 H m−1)

SOLUTION:
Given : = 0.25 , = 100 , = 2.7 × 10−2 , 0 =
4 × 10−7 −1

250

Non Face-to-face SLT

SOLUTION:

a. The cross-sectional area of the solenoid is given by

Hence the self-inductance of the solenoid is

b.

251

Non Face-to-face SLT

CHAPTER 5.5 MUTUAL INDUCTANCE

LO 5.5 Exercise 105

Two coils, X & Y are magnetically coupled. The emf
induced in coil Y is 2.5 V when the current flowing
through coil X increases at the rate of 5 A s-1.
Determine:
(a) the mutual inductance of the coils
(b) the emf induced in coil X if there is a current

flowing through coil Y which increases at the
rate of 1.5 A s-1.