PROBLEM-SOLVING (Mutual Inductance)

EASY WAY TO SOLVE
PROBLEM IN

MUTUAL INDUCTANCE

3-steps to solve problem in physics

1 List ALL the value given
in SI units.

2 Choose a SUITABLE
formula.

3 Substitute the value
into the FORMULA.

Q1

What is the definition mutual induction?

Solution

Mutual induction is a process of
producing an induced E.M.F in one coil
due to the change of current in another
coil.

Q2

Two coils were located side by side. The primary coil
has 1000 turns with 3 cm diameter and 50 cm length.
If the secondary coil has 50 turns, calculate ;
(a) mutual inductance
(b) the induced e.m.f in the secondary coil if the
current flowing in the primary coil is changing at the
rate of 4.8 A s-1.

Example for Solution

(a) 1. Given:

1 = 0.5 m = 2
1
1 = 0.03 m  = 4 (0.03)2
2 = 1000 turns =
= 50 turns 4 10−4 m2
7.07 ×

2. Formula:

= 0 1 2


3. Substitute :

= 0 1 2


= (4 × 10−7)(1000)(50)(7.07 × 10−4 2)
0.5

= 8.88 × 10−5 H

Example for Solution

(b) 1. Given :

1 = 4.8 −1

= 8.88 × 10−5 H
(From previous question)

2. Formula: 1

2 =

3. Substitute :

2 = 1


= 4.26 × 10−4 V

Q3

If the flux in coil 2 is 5 × 10−3 Wb and it
has 300 turns and the current in coil 1 is
7 A.

Calculate the mutual inductance.

Example for Solution

1. Given :

2 = 5 × 10−3Wb
2 = 300 turns

1= 7 A

2. Formula: 2 2
1
=

3. Substitute :

= 2 2
1
(300)(5 × 10−3)
=7

= 0.214 H

Q4

The flux in coil 1 is 3 Wb and it
has x turns and the current in
coil 2 is 2 A, calculate the value
of x if the mutual inductance is
750 H.

Example for Solution

1. Given :

1 = 3
2 = 2
= 750

2. Formula: 1 1
2
=

3. Substitute :

= 1 1
1 = 2 2
1
(750)(2)
= (3)

= 500 turns

Q5

Two coils are placed side by side and fixed
to their positions. When there is no current
in coil 1, the current in coil 2 increases at a
rate of 12 A s-1. The induced e.m.f in coil 1 is
100 mV.
But if there is no current in coil 2 and current
in coil 1 changes at rate of 2.4 A s-1, what
will the induced e.m.f in coil 2?

Example for Solution

1. Given :

Case 1, Case 2,
1 = 0 A
2 2 = 0 A
1
= 12 −1 = 2.4 As−1
× 10−3
1 = 100 V

2. Formula: 1

2 =

3. Substitute : 1
2
=

100 × 10−3
= 12 −1
= 8.33 × 10−3 H

So, 2 = 1

= (8.33 × 10−3)(2.4)
= 20 × 10−3 V

Higher level questions

Q6

Is the principle of dynamically (moving)
induced E.M.F is utilized in transformer?

Solution

NO. The principle of dynamically
(moving) induced E.M.F is utilized in
generator.

Higher level questions

Q7

A long straight solenoid of cross-
sectional area of 10 cm2 and has 10
turns per cm. A short 50 turns coaxial
coil of cross-sectional area 2 cm2 is
fixed inside the solenoid at the middle.
What is the mutual inductance of the
solenoid and the coil is?

Example for Solution

1. Given :

= 10 × 10−4 m2

= 1000 turns m−1

= 50 turns m2
= 2 × 10−4

2. Formula: 0 1 2 2

=

3. Substitute :

= 0 1 2 2

1
= 0 2 2

= 4 × 10−7 (1000)(50)(10 × 10−4)

= 20 H