JEE MainPhysics

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Mutual Inductance of Concentric Coplanar Circular Coils

Q: Why do we assume the magnetic field of the outer coil is uniform over the inner coil?

Because r_1 = 1cm is 1000 times smaller than r_2 = 1000cm, the magnetic field variation across the small area π r_1^2 is negligible.

Two concentric circular coils with radii 1cm and 1000cm, and number of turns 10 and 200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be ______× 10^(-8)H. (Take, π^2=10)

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Step-by-step solutionView

Correct answer

The mutual inductance of the concentric coil system is 4 × 10⁻⁸ H, giving a numerical value of 4.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Inner coil: radius $r_{1} = 1 cm = 10^{- 2} m$ , turns $N_{1} = 10$ . Outer coil: radius $r_{2} = 1000 cm = 10 m$ , turns $N_{2} = 200$ . $μ_{0} = 4 π \times 10^{- 7} H/m$ , $π^{2} = 10$ .

02Find

The mutual inductance $M$ of the arrangement in the form $k \times 10^{- 8} H$ .

03Visualise

Two coplanar, concentric circular loops. Since $r_{1} ≪ r_{2}$ , the magnetic field produced by current $I_{2}$ in the outer coil is approximately uniform across the entire area of the small inner coil and perpendicular to its plane.

04Strategise

Current $I_{2}$ in outer coil produces $B_{2} = \frac{μ_{0} N_{2} I_{2}}{2 r_{2}}$ at the center. Flux linked with the inner coil is $Φ_{1} = N_{1} (B_{2} A_{1}) = N_{1} (\frac{μ_{0} N_{2} I_{2}}{2 r_{2}}) (π r_{1}^{2})$ . By definition $Φ_{1} = M I_{2}$ , hence $M = \frac{μ_{0} N_{1} N_{2} π r_{1}^{2}}{2 r_{2}}$ .

05Execute

Substitute values into the formula: $M = \frac{(4 π \times 10^{- 7}) (10) (200) (π) (10^{- 2})^{2}}{2 \times 10} = \frac{4 π^{2} \times 10^{- 7} \times 2000 \times 10^{- 4}}{20} = \frac{4 (10) \times 2000 \times 10^{- 11}}{20} = 4 \times 10^{- 8} H$ . Thus, the integer value required is 4.

✓Verify

Dimensionally, $[M] = [μ_{0}] \cdot [L] = (H/m) \cdot m = H$ . The ratio $r_{1} / r_{2} = 10^{- 3} ≪ 1$ , confirming the central-field approximation is highly accurate.

Hints that build this answer step by step

Which formula gives the mutual inductance M between two concentric, coplanar circular coils when the inner coil radius r₁ is much smaller than the outer coil radius r₂?
M = (μ₀ · π · r₁² · N₁ · N₂) / (2 · r₂)
What are the given values converted properly into standard SI units?
r₁ = 0.01 m, r₂ = 10 m, N₁ = 10, N₂ = 200
Substituting μ₀ = 4π × 10⁻⁷ H/m and π² = 10 into M = (μ₀ · π · r₁² · N₁ · N₂) / (2 · r₂), what is the calculated value of M?
4 × 10⁻⁸ H

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Physics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why do we assume the magnetic field of the outer coil is uniform over the inner coil?

Because $r_{1} = 1 cm$ is $1000$ times smaller than $r_{2} = 1000 cm$ , the magnetic field variation across the small area $π r_{1}^{2}$ is negligible.

Answer

Accepted answer: 4

The mutual inductance of the concentric coil system is 4 × 10⁻⁸ H, giving a numerical value of 4.

Why each option works or fails

Step-by-step solution

Given: Inner coil: radius $r_{1} = 1 cm = 10^{- 2} m$ , turns $N_{1} = 10$ . Outer coil: radius $r_{2} = 1000 cm = 10 m$ , turns $N_{2} = 200$ . $μ_{0} = 4 π \times 10^{- 7} H/m$ , $π^{2} = 10$ .
Find: The mutual inductance $M$ of the arrangement in the form $k \times 10^{- 8} H$ .
Visualise: Two coplanar, concentric circular loops. Since $r_{1} ≪ r_{2}$ , the magnetic field produced by current $I_{2}$ in the outer coil is approximately uniform across the entire area of the small inner coil and perpendicular to its plane.
Strategise: Current $I_{2}$ in outer coil produces $B_{2} = \frac{μ_{0} N_{2} I_{2}}{2 r_{2}}$ at the center. Flux linked with the inner coil is $Φ_{1} = N_{1} (B_{2} A_{1}) = N_{1} (\frac{μ_{0} N_{2} I_{2}}{2 r_{2}}) (π r_{1}^{2})$ . By definition $Φ_{1} = M I_{2}$ , hence $M = \frac{μ_{0} N_{1} N_{2} π r_{1}^{2}}{2 r_{2}}$ .
Execute: Substitute values into the formula: $M = \frac{(4 π \times 10^{- 7}) (10) (200) (π) (10^{- 2})^{2}}{2 \times 10} = \frac{4 π^{2} \times 10^{- 7} \times 2000 \times 10^{- 4}}{20} = \frac{4 (10) \times 2000 \times 10^{- 11}}{20} = 4 \times 10^{- 8} H$ . Thus, the integer value required is 4.
Verify: Dimensionally, $[M] = [μ_{0}] \cdot [L] = (H/m) \cdot m = H$ . The ratio $r_{1} / r_{2} = 10^{- 3} ≪ 1$ , confirming the central-field approximation is highly accurate.

Mutual Inductance of Concentric Coplanar Circular Coils

What feels right?

The mutual inductance of the concentric coil system is 4 × 10⁻⁸ H, giving a numerical value of 4.

Why each option works or fails

Solution

01Given

02Find

03Visualise

04Strategise

05Execute

✓Verify

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01Given

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03Visualise

04Strategise

05Execute

✓Verify

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