Question 248: Electromagnetic Induction

Question

Two concentric circular coils having radii ' $r_1{ }^{\prime}$ and ' $r_2$ ' $\left(r_2 \ll r_1\right)$ are placed co-axially with centres coinciding. The mutual inductance of the arrangement is ( $\mu_0=$ permeability of free space) (Both coils have single turn)

Accepted Answer

The mutual inductance between two concentric circular coils can be derived by considering the magnetic flux through one coil due to the current in the other coil. For simplicity, consider the larger coil of radius $r_1$ to be producing a magnetic field, while the smaller coil of radius $r_2$ experiences this field.
The magnetic field at the center of the larger coil when a current $I$ flows through it is given by:
$$ B_1 = \frac{\mu_0 I}{2 r_1} $$
where $B_1$ is the magnetic field at the center, $\mu_0$ is the permeability of free space, and $I$ is the current flowing in the larger coil.
The magnetic flux $\Phi_2$ through the smaller coil (of radius $r_2$) is given by the product of the magnetic field produced by the first coil and the area of the second coil:
$$ \Phi_2 = B_1 \cdot \pi r_2^2 = \frac{\mu_0 I}{2 r_1} \cdot \pi r_2^2 $$
Mutual inductance $M$ is defined as the ratio of the magnetic flux through one coil to the current in the other coil:
$$ M = \frac{\Phi_2}{I} = \frac{\mu_0 \pi r_2^2}{2 r_1} $$
Thus, the correct option for the mutual inductance of the arrangement is:
Option D
$$\frac{\mu_0 \pi r_2^2}{2 r_1}$$

Answer

$\frac{\mu_0 \pi r_2}{2 r_1}$

Answer

$\frac{\mu_0 \pi}{2 r_1 r_2}$

Answer

$\frac{\mu_0 \pi r_1}{2 r_2}$

Answer

$\frac{\mu_0 \pi r_2^2}{2 r_1}$

Electromagnetic Induction

Two concentric circular coils having radii ' r_1{ }^{\prime} and ' r_2 ' \left(r_2 \ll r_1\right) are placed co-axially with centres coinciding. The mutual inductance of the arrangement is ( \mu_0= permeability of free space) (Both coils have single turn)

Choose the correct answer: