EASY WAY TO
SOLVE PROBLEM IN
Induced E.M.F
3-Steps to solve problem in physics
GIVEN FORMULA APPLY
1 2 3
List ALL the Choose a Substitute the
value given in SI suitable value into the
FORMULA.
units. formula.
Q1
The working principle of
transformer depends on;
a) Coulomb’s Law
b) Faraday’s Law
c) Ampere’s Law
d) Lenz’s Law
Solution
The working principle of
transformer depends on;
a) Coulomb’s Law
b) Faraday’s Law
c) Ampere’s Law
d) Lenz’s Law
Q2
If current in a conductor
increases, self-induced E.M.F
will produce current opposite
to the increasing current. Is
the statement true or false?
Solution
TRUE. According to Lenz’s Law,
the direction of induced current
will always in the direction that
opposes the change in the
magnetic flux that causes it.
× × × ×A × × × × × ×
Q3 × × × × × × × × × ×
×× ××××××××
×× ××××××××
× × × ×B × × × × × ×
A circular coil has 200 turns with a diameter
of 36 cm and a resistance of 2.0 Ω.
A uniform magnetic field is applied
perpendicularly to the plane of the coil and
changes uniformly from 0.5 T to 0 T in 0.8 s.
Q3
a. Find the induced E.M.F and
current in the coil while the field
changed.
b. Determine the direction of the
current induced.
Example of Solution
(a)1. Given :
= 200 turns
= 36 cm 36 × 10−2 m
= 2 Ω
= 0 T
= 0.5
∆ = 0.8
2. Formula :
=
= −
( * N and A constant,
θ = 0° )
= − −
Δ
3. Apply :
Area of the coil, A
Thus, the induced E.M.F is
= − −
= Δ
(0 − 0.5)
−(200)(0.1018) 0.8
= 12.725 V#
To determine the value of induced current,
Use: =
=
12.725
=2
= 6.36 A #
(b) 1. Given :
X : B is into the page
2. Formula :
Lens Law - Induced current will try
to prevent the decrease in flux in
order to maintain the magnetic field
lines passing through the coil.
3. Apply : I induced
Use
Right Hand Rule.
Thumb – In direction of external B.
Curl Finger – Show the direction of
induced current.
Thus, induced current is in clockwise direction.
#
Axis of rotation
Q4
A loop of area 0.10 m2 is rotating at 60 rev s-1
with its axis of rotation perpendicular to a
0.20 T magnetic field.
Q4
a. If the loop have 1000 turns, what is the
maximum voltage induced in the loop?
b. What is the orientation of the loop with
respect to the magnetic field at maximum
induced voltage?
Example for solution
(a) 1. Given :
= 0.1 m2
= 0.2 T
= 60 rev s − 1 =
= 120 rad s − 1
= 1000 turns
2. Formula :
=
Note for maximum value, sin = 1.0
3. Apply :
Thus, maximum voltage induced is
=
= (1000)(0.2)(0.1)(120 )
= 7.54 × 103 V #
(b) 1. Given;
Axis of rotation PERPENDICULAR to
magnetic field, B
2. Formula :
=
3. Apply:
Maximum value, sin = 1.0
= sin
Thus, the orientation of the loop must be parallel
to the field #
Q5 R=6Ω L = 1.2m
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vx x x x x x x x x x x x x x
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Figure 3
Based on Figure 3, assume that a
uniform 2.50 T magnetic field is
directed into the page.
Q5
a. At what speed should the blue bar
be moved to produced a current of 0.5
A in the resistor?
b. What is the direction of the
induced current ?
Example of solution
(a) 1. Given;
= 6 Ω
= 1.2 m
= 2.50 T
= 0.5 A
= ?
2. Formula :
= sin
3. Apply :
= sin
Thus, the speed of the bar is
= sin (0.5)(6)
= (2.5)(1.2) sin(90)
= 1 ms−1
#
(b) 1. Given : V
: into the page B
= right I
2. Formula:
Fleming’s Right Hand
Rule
3. Apply :
xx xx xx xx xx xx xx
xx xx xx xx xx xx xx
xx xx xx xx xx
xx xx x x vx x
xx x x R =x6 Ωx xx xL =x1.2m
xx xx xx xx
xx xx xx xx xx xx xx
xx xx xx xx xx
xx xx xx xx xx
Thus the direction of the induced
current is anticlockwise. #
Q6
A single turn of wire of cross-
sectional area 5.0 cm2 is at 90o to
a magnetic field of 0.02 T, which
is reduced to 0 in 10 s at a steady
rate. What is the E.M.F.
induced?
Example of solution
1. Given : = 5.0 cm2
= 0°
= 0 T
= 0.02 T
Δ = 10 s
= 1 turns
2. Formula :
= − ( )
3. Apply:
Thus, the induced E.M.F is
= −(1) 5.0 × 10−4 0.02
cos 0° 10
= −1.0 × 10−6 V#
× × A× × × × × ×
Q7
×× ××××××
× × B× × × × × ×
AA
The flexible loop has a radius of 12 cm
and is in 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.
Q7
If it takes 0.20 s to close the loop, find
the magnitude of the average induced
E.M.F in it during this time.
× × A× × × × × × ×× ×× ×× ××
×× ×××××× ×× ×× ×× ××
× × B× × × × × × ×× ×× ×× ××
Example of solution
1. Given :
= 0.12 m
∆ = 0.2 s
= 0.15 T
= 0 T
2. Formula :
= −
3. Apply:
Thus, induced E.M.F is
= − −
Δ
(0.12)2)
= −(0.15) (0 − 0.2
= 3.4 × 10−2 V #
Higher level question Q
Q8
P
Two plane coils P and Q are arranged coaxially as
shown. Using an external battery, an increasing
or decreasing clockwise current can be made to
flow through coil P. What is the direction of the
induced current through coil Q if the current in
coil P is increase? And what if it decrease?
Solution
Q
PI
1. Given :
The current in coil P is in clockwise direction.
2. Formula/concept :
Lenz’s Law concept - induced current always
oppose the change in the magnetic flux that
causes it.
Solution
3. Apply:
P I
Figure (a): If P INCREASE,
induced current in coil Q will N
move in ANTICLOCKWISE
Figure (a)
direction because it oppose
the increasing magnetic flux
from coil P. Figure (b): If P INCREASE,
induced current in coil Q
PI will move in
ANTICLOCKWISE direction
N because it oppose the
increasing magnetic flux
Figure (b) from coil P.
Higher level question
Q9
Select the statement below that affect the
strength of the induced current.
I. The strength of the magnet
II. The size of the magnet
III. How quickly the flux lines cut by the
moving magnet
IV. The type of the wire
Solution
I and III
The strength of the magnet (MAGNETIC
FIELD STRENGTH) and how quickly the
flux lines cut by the moving magnet
(VELOCITY).