1. Introduction
When two coils or loops are placed near each other, a changing current in
one loop can produce magnetic flux through the other loop.
The corresponding induced EMF is described using
mutual inductance.
M = Φ₂₁ / I₁
where:
- M = mutual inductance
- Φ₂₁ = magnetic flux through coil 2 due to current I₁ in coil 1
- I₁ = current in coil 1
2. Mutual Inductance
If current in coil 1 changes, the induced EMF in coil 2 is:
ε₂ = -M dI₁/dt
The negative sign represents
Lenz's law.
Important:
Mutual inductance depends on the geometry, relative orientation,
distance and number of turns of the two circuits.
3. Large Loop and Tiny Loop
Consider a very large square loop carrying current I.
A tiny loop of area A is placed near its centre.
If the magnetic field produced by the large loop is approximately uniform
over the small loop, then:
Φ = BA cosθ
For the small loop whose plane is perpendicular to the magnetic field:
Φ = BA
Therefore:
M = Φ/I
Hence:
M = BA/I
4. Magnetic Field at the Centre of a Square Loop
Let the side of the large square loop be a and current I flows through it.
The magnetic field at the centre is obtained by adding the fields due to
all four sides.
For a finite straight conductor:
B = μ₀I/(4πr)(sinθ₁ + sinθ₂)
At the centre of a square:
- Distance from centre to each side = a/2
- Angle made by each side at the centre = 45° on each end
Thus for one side:
B₁ = μ₀I/(4π(a/2))
(sin45° + sin45°)
Since:
sin45° = 1/√2
We get:
B₁ = μ₀I√2/(2πa)
There are four identical sides:
B = 2√2 μ₀I/(πa)
5. Mutual Inductance with a Tiny Loop
Let the small loop have area A and be sufficiently small that the magnetic
field can be considered uniform over it.
The magnetic flux through the small loop is:
Φ = BA
Substituting the field at the centre of the square:
Φ =
[2√2 μ₀I/(πa)]A
Therefore:
M = Φ/I
Hence:
M = 2√2 μ₀A/(πa)
This is the important result for a tiny loop placed at the centre of a
large square loop, with the small loop oriented so that its normal is
along the magnetic field.
6. If the Small Loop is Tilted
If the normal to the small loop makes an angle θ with the magnetic field,
the magnetic flux is:
Φ = BA cosθ
Therefore:
M = BA cosθ / I
For the square loop:
M =
2√2 μ₀A cosθ/(πa)
7. Non-Coplanar Arrangement
When the small loop is not in the same plane as the large loop, the mutual
inductance depends on the angle between the magnetic field at the small
loop and the normal to the small loop.
M = Φ/I
If the field is approximately uniform across the small loop:
M = BA cosθ/I
Therefore:
M = BAI⁻¹ cosθ
8. Effect of Distance
The approximation:
Φ ≈ BA
is valid when the small loop is sufficiently small and lies in a region
where the magnetic field does not change appreciably.
If the small loop is moved away from the centre, the magnetic field at
its position changes and the simple centre-field expression may no
longer be used directly.
For a tiny loop:
M ≈ B(position) A cosθ / I
9. Reciprocity of Mutual Inductance
For two passive circuits in a linear medium:
M₁₂ = M₂₁ = M
This means mutual inductance of coil 1 with coil 2 is equal to that of
coil 2 with coil 1.
10. SI Unit
The SI unit of mutual inductance is:
Henry (H)
One henry is the mutual inductance when a current changing at 1 A/s in
one circuit produces an induced EMF of 1 V in the other circuit.
11. Solved Example 1
A square loop of side 2 m carries a current of 5 A.
A tiny loop of area 0.01 m² is placed at its centre with its normal
along the magnetic field. Find the mutual inductance.
For a square:
B = 2√2 μ₀I/(πa)
Mutual inductance:
M = 2√2 μ₀A/(πa)
Putting:
a=2 m, A=0.01 m²
M =
2√2(4π×10⁻⁷)(0.01)/(π×2)
M ≈ 5.66 × 10⁻⁹ H
12. Solved Example 2 — Tilted Loop
For the same arrangement, if the normal of the small loop makes an angle
60° with the magnetic field, find the new mutual inductance.
M = M₀ cos60°
Since:
cos60° = 1/2
Therefore:
M = M₀/2
13. Solved Example 3 — Induced EMF
A tiny loop has mutual inductance M = 2 mH with a large loop.
If current in the large loop changes at 500 A/s, find the induced EMF.
ε = -M dI/dt
Magnitude:
|ε|=(2×10⁻³)(500)
|ε| = 1 V
14. Solved Example 4 — Zero Mutual Inductance
If the normal of the small loop is perpendicular to the magnetic field,
then θ=90°.
M = BA cos90°/I
Since:
cos90°=0
Therefore:
M = 0
15. Important Formula Sheet
| Quantity |
Formula |
| Mutual inductance |
M=Φ/I |
| Induced EMF |
ε=-M dI/dt |
| Flux |
Φ=BA cosθ |
| Small loop |
M=BA cosθ/I |
| Centre field of square |
B=2√2 μ₀I/(πa) |
| Square + tiny loop |
M=2√2 μ₀A cosθ/(πa) |
| SI unit |
Henry (H) |
| Reciprocity |
M₁₂=M₂₁ |
Section A — 1 Mark MCQs
10 Questions
Q1. The SI unit of mutual inductance is:
1 Mark
A) Tesla
B) Weber
C) Henry
D) Ohm
Answer: C
Q2. Mutual inductance is defined as:
1 Mark
A) I/Φ
B) Φ/I
C) ΦI
D) I²/Φ
Answer: B
Q3. The induced EMF due to mutual induction is:
1 Mark
A) M/I
B) MI
C) -M dI/dt
D) M dI/dt²
Answer: C
Q4. For a small loop in a uniform magnetic field, flux is:
1 Mark
A) BA
B) BA cosθ
C) B/A
D) A/B
Answer: B
Q5. If the normal of the small loop is parallel to B, θ is:
1 Mark
A) 0°
B) 30°
C) 60°
D) 90°
Answer: A
Q6. If θ=90°, mutual inductance is:
1 Mark
A) Maximum
B) Zero
C) Infinite
D) Negative always
Answer: B
Q7. Mutual inductance of two passive circuits satisfies:
1 Mark
A) M₁₂ = -M₂₁
B) M₁₂ = M₂₁
C) M₁₂ = 0 always
D) M₁₂ = 2M₂₁
Answer: B
Q8. The magnetic field at the centre of a square loop is proportional to:
1 Mark
A) I/a
B) Ia
C) a/I
D) I/a²
Answer: A
Q9. Mutual inductance increases if the area of the small loop:
1 Mark
A) Increases
B) Decreases
C) Becomes zero
D) Has no effect
Answer: A
Q10. Lenz's law determines the:
1 Mark
A) Magnitude only
B) Direction of induced current
C) Resistance
D) Area
Answer: B
Section B — 2 Marks
10 Questions
Q11. Define mutual inductance.
2 Marks
Mutual inductance is the flux linked with one circuit per unit current
in the other circuit:
M=Φ/I
Q12. Write the relation between induced EMF and mutual inductance.
2 Marks
Q13. What factors affect mutual inductance?
2 Marks
It depends on the geometry, size, relative position, orientation,
number of turns and magnetic properties of the medium.
Q14. What happens to M when the small loop is tilted through 60°?
2 Marks
Q15. Why can the magnetic field be considered uniform over a tiny loop?
2 Marks
Because the dimensions of the tiny loop are much smaller than the scale
over which the magnetic field changes appreciably.
Q16. State the reciprocity relation for mutual inductance.
2 Marks
Q17. What is the mutual inductance if the magnetic flux through the
second loop is zero?
2 Marks
M=Φ/I=0
Therefore mutual inductance is zero.
Q18. Write the magnetic field at the centre of a square loop of side a.
2 Marks
Q19. What happens to mutual inductance if the area of the small loop is
doubled?
2 Marks
Since M is directly proportional to A:
M∝A
Doubling A doubles M, under the small-loop approximation.
Q20. Why does a changing current produce induced EMF in the second loop?
2 Marks
A changing current produces a changing magnetic field. This changes the
magnetic flux through the second loop and hence induces EMF according
to Faraday's law.
Section C — 3 Marks
10 CBSE + Competitive Questions
Q21. Derive the mutual inductance between a large square loop and a tiny
loop at its centre.
3 Marks
For a square loop of side a:
B=2√2 μ₀I/(πa)
Flux through tiny loop of area A:
Φ=BA
Therefore:
M=Φ/I
Hence:
M=2√2 μ₀A/(πa)
Q22. Derive the mutual inductance when the tiny loop is tilted at angle θ.
3 Marks
Flux:
Φ=BAcosθ
Therefore:
M=BAcosθ/I
For a square loop:
M=2√2 μ₀A cosθ/(πa)
Q23. A square loop of side 1 m carries current I. Find its magnetic
field at the centre.
3 Marks
B=2√2 μ₀I/(πa)
For a=1 m:
B=2√2 μ₀I/π
Q24. A tiny loop of area 2×10⁻³ m² is placed at the centre of a square
loop of side 2 m. Find M.
3 Marks
M=2√2 μ₀A/(πa)
Putting values:
M=
2√2(4π×10⁻⁷)(2×10⁻³)/(π×2)
M≈1.13×10⁻⁹ H
Q25. If mutual inductance is 5 mH and current changes at 200 A/s,
find induced EMF.
3 Marks
|ε|=M dI/dt
|ε|=(5×10⁻³)(200)
|ε|=1 V
Q26. Explain why mutual inductance becomes zero when the normal of the
small loop is perpendicular to B.
3 Marks
Flux:
Φ=BAcosθ
For θ=90°:
Φ=BAcos90°=0
Therefore:
M=Φ/I=0
Q27. A small loop is rotated from θ=0° to θ=60°. By what fraction does
its mutual inductance change?
3 Marks
Initially:
M₀=BA/I
Finally:
M=M₀cos60°=M₀/2
Therefore the mutual inductance becomes half its initial value.
Decrease = 50%
Q28. A large square loop has side 4 m and carries 10 A current.
A tiny loop of area 0.02 m² is at its centre. Find the flux through
the tiny loop.
Take μ₀=4π×10⁻⁷ H/m.
3 Marks
Field at centre:
B=2√2 μ₀I/(πa)
B=
2√2(4π×10⁻⁷)(10)/(π×4)
B≈2.83×10⁻⁶ T
Flux:
Φ=BA
Φ=(2.83×10⁻⁶)(0.02)
Φ≈5.66×10⁻⁸ Wb
Q29. Show that the mutual inductance between a square loop and a tiny
loop is independent of the current in the large loop.
3 Marks
Magnetic field at centre:
B=2√2 μ₀I/(πa)
Flux:
Φ=BA
Therefore:
Φ=
2√2 μ₀IA/(πa)
Mutual inductance:
M=Φ/I
Thus:
M=2√2 μ₀A/(πa)
The current I cancels out.
Q30. Explain the physical meaning of mutual inductance and state two
factors affecting it.
3 Marks
Mutual inductance measures how effectively a changing current in one
circuit produces magnetic flux and induced EMF in another circuit.
M=Φ/I
Two important factors are:
- Relative position and orientation of the loops.
- Size/geometry and number of turns of the loops.
The magnetic properties of the medium can also affect mutual inductance.
16. Quick Revision
Remember these results:
M=Φ/I
ε=-M dI/dt
Φ=BAcosθ
Bsquare=2√2 μ₀I/(πa)
Msquare-small
=
2√2 μ₀A cosθ/(πa)
M₁₂=M₂₁
Special cases:
θ=0° → M is maximum.
θ=90° → M=0.
If A doubles → M doubles.
If side a of the large square increases → M decreases as 1/a,
within the centre-field/tiny-loop approximation.
