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NVAMediumJEE 2023Self & Mutual Inductance

JEE Physics 2023 Question with Solution

Two concentric circular coils with radii 1 cm1 \, \text{cm} and 1000 cm1000 \, \text{cm}, and number of turns 1010 and 200200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be _____ ×108\times 10^8 H.

Answer

Correct answer:4

Step-by-step solution

Standard Method

Given: Two concentric circular coils are coaxial with radii a=1000 cma = 1000 \, \text{cm} and b=1 cmb = 1 \, \text{cm}, and number of turns N=200N = 200 and n=10n = 10.

Find: The mutual inductance of the arrangement.

Two concentric circular coils with common center, the larger coil labeled N and radius a, and the smaller inner coil labeled n and radius b.

As the larger coil is taken as primary,

M=μ0Nnb22aM = \frac{\mu_0 N n b^2}{2a}

Substituting the given values,

M=4π×10−7×200×10×π×1×10−42×1000×10−2M = \frac{4\pi \times 10^{-7} \times 200 \times 10 \times \pi \times 1 \times 10^{-4}}{2 \times 1000 \times 10^{-2}}

Thus,

M=4×10−8 HM = 4 \times 10^{-8} \, \text{H}

Therefore, the value of mutual inductance is 4×10−8 H4 \times 10^{-8} \, \text{H}.

Using the field of the larger coil

Given: The larger coil has radius aa and turns NN, while the smaller coil has radius bb and turns nn.

Find: Mutual inductance between the two coils.

For the larger circular coil, the magnetic field at its center is used, and the flux through the smaller coil is obtained from that field over the smaller area.

The area of the smaller coil is

A=πb2A = \pi b^2

and the mutual inductance expression used in the solution is

M=μ0Nnb22aM = \frac{\mu_0 N n b^2}{2a}

On substitution of the numerical values given in the solution,

M=4×10−8 HM = 4 \times 10^{-8} \, \text{H}

Hence, the required numerical value is 44, and the mutual inductance is 4×10−8 H4 \times 10^{-8} \, \text{H}.

Common mistakes

  • Using the radius of the larger coil in the flux area is incorrect because the flux is linked through the smaller coil. Use the area corresponding to πb2\pi b^2, not the larger circular area.

  • Failing to convert cm to m gives the wrong power of 1010. Convert 1000 cm=10 m1000 \, \text{cm} = 10 \, \text{m} and 1 cm=10−2 m1 \, \text{cm} = 10^{-2} \, \text{m} before substitution.

  • Interchanging primary and secondary without checking the formula can lead to an inconsistent substitution. Follow the stated relation carefully with the larger coil taken as primary.

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