A small square loop of wire of side l is placed inside a large square loop of wire L(L\gt >l) . Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is :

Electromagnetic Induction PYQ — JEE Main 2022 · 25 Jul · Shift 1 · Q53 with Solution

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Correct answer: C

Goal

We need the mutual inductance $M$ between:

a large square loop of side $L$
a small square loop of side $l$

with $L \gg l$ , both coplanar and concentric.

Mutual inductance is

M = \frac{\Phi}{I}

where $\Phi$ is the magnetic flux through the small loop due to current $I$ in the large loop.

Use the approximation $L \gg l$

Since the small loop is very small compared to the large loop, the magnetic field due to the large loop is approximately uniform over the small loop and equal to the field at the common center $O$ .

So,

\Phi = B_O \cdot l^2

and hence

M = \frac{B_O l^2}{I}

Thus we only need the magnetic field at the center of the large square loop.

Magnetic field at the center due to one side of the large square

For a finite straight wire,

B = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2)

At the center of the square:

perpendicular distance from center to any side is

r = \frac{L}{2}

the center sees each side symmetrically, so

\theta_1 = \theta_2 = 45^\circ

Therefore field due to one side is

B_1 = \frac{\mu_0 I}{4\pi (L/2)}(\sin45^\circ + \sin45^\circ)

Since $\sin45^\circ = \frac{1}{\sqrt2}$ ,

B_1 = \frac{\mu_0 I}{2\pi L}\left(\frac{2}{\sqrt2}\right) = \frac{\mu_0 I}{\sqrt2\,\pi L}

Field due to all four sides

All four sides contribute in the same perpendicular direction, so

B_O = 4B_1 = 4\cdot \frac{\mu_0 I}{\sqrt2\,\pi L} = \frac{4\mu_0 I}{\sqrt2\,\pi L} = \frac{2\sqrt2\,\mu_0 I}{\pi L}

Flux through the small loop

Since field is nearly uniform over the small square,

\Phi = B_O \cdot l^2 = \frac{2\sqrt2\,\mu_0 I}{\pi L} \, l^2

So mutual inductance is

M = \frac{\Phi}{I} = \frac{2\sqrt2\,\mu_0 l^2}{\pi L}

Match with the options

M = \frac{2\sqrt2\,\mu_0 l^2}{\pi L}

which matches Option C.

Verification with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

More from Electromagnetic Induction

View written solutionFree

Correct answer: C

Goal

We need the mutual inductance $M$ between:

a large square loop of side $L$
a small square loop of side $l$

with $L \gg l$ , both coplanar and concentric.

Mutual inductance is

M = \frac{\Phi}{I}

where $\Phi$ is the magnetic flux through the small loop due to current $I$ in the large loop.

Use the approximation $L \gg l$

Since the small loop is very small compared to the large loop, the magnetic field due to the large loop is approximately uniform over the small loop and equal to the field at the common center $O$ .

So,

\Phi = B_O \cdot l^2

and hence

M = \frac{B_O l^2}{I}

Thus we only need the magnetic field at the center of the large square loop.

Magnetic field at the center due to one side of the large square

For a finite straight wire,

B = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2)

At the center of the square:

perpendicular distance from center to any side is

r = \frac{L}{2}

the center sees each side symmetrically, so

\theta_1 = \theta_2 = 45^\circ

Therefore field due to one side is

B_1 = \frac{\mu_0 I}{4\pi (L/2)}(\sin45^\circ + \sin45^\circ)

Since $\sin45^\circ = \frac{1}{\sqrt2}$ ,

B_1 = \frac{\mu_0 I}{2\pi L}\left(\frac{2}{\sqrt2}\right) = \frac{\mu_0 I}{\sqrt2\,\pi L}

Field due to all four sides

All four sides contribute in the same perpendicular direction, so

B_O = 4B_1 = 4\cdot \frac{\mu_0 I}{\sqrt2\,\pi L} = \frac{4\mu_0 I}{\sqrt2\,\pi L} = \frac{2\sqrt2\,\mu_0 I}{\pi L}

Flux through the small loop

Since field is nearly uniform over the small square,

\Phi = B_O \cdot l^2 = \frac{2\sqrt2\,\mu_0 I}{\pi L} \, l^2

So mutual inductance is

M = \frac{\Phi}{I} = \frac{2\sqrt2\,\mu_0 l^2}{\pi L}

Match with the options

M = \frac{2\sqrt2\,\mu_0 l^2}{\pi L}

which matches Option C.

Verification with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

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