Two conducting circular loops A and B are placed in the same plane with their centres coinciding as shown in figure. The mutual inductance between them is:

\frac\mu_o2 \pi \cdot \fracb^2a

\frac\mu_02 \pi \cdot \fraca^2b

Electromagnetic Induction PYQ — JEE Main 2024 · 5 Apr · Shift 1 · Q70 with Solution

View written solutionFree

Correct answer: B

Interpret the geometry

Two circular conducting loops $A$ and $B$ are coplanar and concentric.

Let the radii be:
- loop $A$ : radius $a$
- loop $B$ : radius $b$
From the options, it is implied that one loop is much larger than the other, and the mutual inductance is to be found using the magnetic field at the center due to one loop, assumed nearly uniform over the smaller loop.
Magnetic field due to a circular loop

The magnetic field at the center of a circular loop of radius $b$ carrying current $I$ is
$B = \frac{\mu_0 I}{2b}.$
Flux linked with the smaller loop

If the larger loop has radius $b$ and the smaller loop has radius $a$ , then the field produced by the larger loop may be taken approximately constant over the area of the smaller loop.

Area of smaller loop:
$A = \pi a^2.$
Hence magnetic flux through the smaller loop is
$\Phi = BA = \frac{\mu_0 I}{2b} \cdot \pi a^2.$
Definition of mutual inductance

Mutual inductance $M$ is given by
$M = \frac{\Phi}{I}.$
Therefore,
$M = \frac{1}{I}\left(\frac{\mu_0 I}{2b} \cdot \pi a^2\right) = \frac{\mu_0 \pi a^2}{2b}.$
Match with the options
$M = \frac{\mu_0 \pi a^2}{2b}$
which corresponds to Option B.
Verification with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.

View written solutionFree

Correct answer: B

Interpret the geometry

Two circular conducting loops $A$ and $B$ are coplanar and concentric.

Let the radii be:
- loop $A$ : radius $a$
- loop $B$ : radius $b$
From the options, it is implied that one loop is much larger than the other, and the mutual inductance is to be found using the magnetic field at the center due to one loop, assumed nearly uniform over the smaller loop.
Magnetic field due to a circular loop

The magnetic field at the center of a circular loop of radius $b$ carrying current $I$ is
$B = \frac{\mu_0 I}{2b}.$
Flux linked with the smaller loop

If the larger loop has radius $b$ and the smaller loop has radius $a$ , then the field produced by the larger loop may be taken approximately constant over the area of the smaller loop.

Area of smaller loop:
$A = \pi a^2.$
Hence magnetic flux through the smaller loop is
$\Phi = BA = \frac{\mu_0 I}{2b} \cdot \pi a^2.$
Definition of mutual inductance

Mutual inductance $M$ is given by
$M = \frac{\Phi}{I}.$
Therefore,
$M = \frac{1}{I}\left(\frac{\mu_0 I}{2b} \cdot \pi a^2\right) = \frac{\mu_0 \pi a^2}{2b}.$
Match with the options
$M = \frac{\mu_0 \pi a^2}{2b}$
which corresponds to Option B.
Verification with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.