Question 232: Electromagnetic Induction

Question

Two conducting circular loops of radii $R_1$ and $\mathrm{R}_2$ are placed in the same plane with their centres coinciding. If $R_1>R_2$, the mutual inductance M between them will be directly proportional to

Accepted Answer

Let us analyze the situation step by step:
Understanding Mutual Inductance ($M$):
Mutual inductance between two coils is given by
$$      M = \frac{\Phi_{21}}{I_1}      $$
where $\Phi_{21}$ is the magnetic flux linked with the second coil due to the current $I_1$ in the first coil.
Loops Arrangement:
Both loops are concentric and coplanar, with $R_1 > R_2$.
Field due to Large Loop ($R_1$) at its Centre:
Magnetic field at the centre of a loop of radius $R_1$ carrying current $I_1$:
$$      B = \frac{\mu_0 I_1}{2 R_1}      $$
Flux through the Smaller Loop ($R_2$):
The flux through the smaller loop is:
$$      \Phi_{21} = B \cdot 	ext{Area of smaller loop}      $$
$$      \Phi_{21} = \left(\frac{\mu_0 I_1}{2 R_1}ight) \pi R_2^2      $$
Calculating $M$:
$$    M = \frac{\Phi_{21}}{I_1} = \frac{\mu_0}{2 R_1} \pi R_2^2    $$
So, $M \propto \frac{R_2^2}{R_1}$
Final Answer:
$\boxed{\dfrac{R_2^2}{R_1}}$
So, the correct option is
Option D
.

Answer

$\frac{\mathrm{R}_1}{\mathrm{R}_2}$

Answer

$\frac{R_2}{R_1}$

Answer

$\frac{\mathrm{R}_1^2}{\mathrm{R}_2}$

Answer

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

Electromagnetic Induction

Two conducting circular loops of radii R_1 and \mathrm{R}_2 are placed in the same plane with their centres coinciding. If R_1>R_2, the mutual inductance M between them will be directly proportional to

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