Q.43. A circular current loop of radius is placed inside square loop of side length such that they

Q: What is the correct answer for this exam Physics Q43?

The correct answer is D. The mutual inductance $M$ between two coils is defined as $M = \frac{\Phi_2}{I_1}$, where $\Phi_2$ is the magnetic flux through the second coil due to current $I_1$ in the first coil. Here, let the current $I$ flow in the square loop. Since $L \gg R$, the magnetic field produced by the square loop can be considered approximately uniform over the area of the circular loop. The magnetic field at the

Q: Q.43. A circular current loop of radius is placed inside square loop of side length such that they are co-planar and their centers coincide. The permeability of free space is . The mutual…

The correct answer is D. The mutual inductance $M$ between two coils is defined as $M = \frac{\Phi_2}{I_1}$, where $\Phi_2$ is the magnetic flux through the second coil due to current $I_1$ in the first coil. Here, let the current $I$ flow in the square loop. Since $L \gg R$, the magnetic field produced by the square loop can be considered approximately uniform over the area of the circular loop. The magnetic field at the

Q: What are the options for this this exam Physics question?

A. $2\sqrt{2}\mu_0 L/R$ · B. $2\sqrt{2}\mu_0 R/L$ · C. $\sqrt{2}\mu_0 R^2/L$ · D. $2\sqrt{2}\mu_0 R^2/L$

Question

Q.43. A circular current loop of radius  is placed inside square loop of side length  such that they are co-planar and their centers coincide. The permeability of free space is . The mutual inductance between circular loop and square loop is __________

FirstInTest · Accepted Answer

The mutual inductance $M$ between two coils is defined as $M = \frac{\Phi_2}{I_1}$, where $\Phi_2$ is the magnetic flux through the second coil due to current $I_1$ in the first coil.
Here, let the current $I$ flow in the square loop. Since $L \gg R$, the magnetic field produced by the square loop can be considered approximately uniform over the area of the circular loop.
The magnetic field at the center of a square loop of side $L$ carrying current $I$ is given by $B = \frac{2\sqrt{2}\mu_0 I}{\pi L}$. This is derived by summing the field from four straight wires. For a single straight wire of length $L/2$ at distance $L/2$ from the center, $B = \frac{\mu_0 I}{4\pi (L/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 I}{2\pi L} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 I}{2\pi L} \frac{2}{\sqrt{2}} = \frac{\mu_0 I}{\sqrt{2}\pi L}$. For four sides, $B_{	ext{center}} = 4 	imes \frac{\mu_0 I}{\sqrt{2}\pi L} = \frac{4\mu_0 I}{\sqrt{2}\pi L} = \frac{2\sqrt{2}\mu_0 I}{\pi L}$.
The magnetic flux through the circular loop (area $A_{	ext{circle}} = \pi R^2$) due to this field is $\Phi = B A_{	ext{circle}}$.
$\Phi = \frac{2\sqrt{2}\mu_0 I}{\pi L} 	imes \pi R^2 = \frac{2\sqrt{2}\mu_0 I R^2}{L}$.
The mutual inductance $M = \frac{\Phi}{I} = \frac{2\sqrt{2}\mu_0 R^2}{L}$.

Answer

$2\sqrt{2}\mu_0 L/R$

Answer

$2\sqrt{2}\mu_0 R/L$

Answer

$\sqrt{2}\mu_0 R^2/L$

Answer

$2\sqrt{2}\mu_0 R^2/L$

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