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Two conducting circular loops of radii \mathrm{R_{1}} and \mathrm{R_{2}} are placed in the same plane with their centres coinciding. If \mathrm{R_{1}> R_{2}} the mutual inductance M between them will be directly proportional to:

Option: 1

\mathrm{\frac{R_1}{R_2}}


Option: 2

\mathrm{\frac{R_2}{R_1}}


Option: 3

\mathrm{\frac{R_1^2}{R_2}}


Option: 4

\mathrm{\frac{R_2^2}{R_1}}


Answers (1)

best_answer

Let a current \mathrm{\mathrm{I}_1} flows through the outer circular coil of radius \mathrm{R}_2 .
The magnetic field at the centre of the coil is

\mathrm{\mathrm{B}_1=\frac{\mu_0 \mathrm{I}_1}{2 \mathrm{R}_1}}

As the inner coil of radius \mathrm{R}_2  placed co-axially has small radius \left(\mathrm{R}_2<\mathrm{R}_1\right), therefore \mathrm{B_{1}} may be taken constant over its cross-sectional area.

Hence, flux associated with the inner coil is \mathrm{\phi_2=\mathrm{B}_1 \pi \mathrm{R}_2^2=\frac{\mu_0 \mathrm{I}_1}{2 \mathrm{R}_1} \pi \mathrm{R}_2^2}

\mathrm{\begin{aligned} & \text { As } M=\frac{\phi_2}{I_1}=\frac{\mu_0 \pi R_2^2}{2 R_1} \\ & \therefore M \propto \frac{R_2^2}{R_1} \end{aligned}}

 

Posted by

Kuldeep Maurya

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