Question 320: Electromagnetic Induction

Question

A circular coil of radius '$$R$$' has '$$N$$' turns of a wire. The coefficient of self induction of the coil will be ( $$\mu_0=$$ permeability of free space)

Accepted Answer

To find the coefficient of self-induction (also known as the inductance) of a circular coil, we need to use the following formula for the inductance of a circular loop:
$$L = \frac{{\mu_0 N^2 A}}{ \ell}$$
Where:
$$L$$ is the inductance
$$\mu_0$$ is the permeability of free space
$$N$$ is the number of turns
$$A$$ is the area of the loop
$$\ell$$ is the mean length of the magnetic flux path
For a circular coil of radius $$R$$, the area $$A$$ is given by
$$A = \pi R^2$$
Since we are dealing with a circular coil, the magnetic flux path would be the circumference of the coil, given by:
$$\ell = 2 \pi R$$
Thus, substituting these values into the equation for inductance, we get:
$$L = \frac{{\mu_0 N^2 \pi R^2}}{2 \pi R}$$
Simplifying the fraction, we find:
$$L = \frac{{\mu_0 N^2 \pi R^2}}{2 \pi R} = \frac{{\mu_0 N^2 R}}{2}$$
So, the coefficient of self-induction of the coil is:
$$\frac{\mu_0 N^2 \pi R}{2}$$
Thus, the correct option is
Option C
.

Answer

$$\frac{\mu_0 N \pi R^2}{2}$$

Answer

$$\frac{\mu_0 N \pi R}{4}$$

Answer

$$\frac{\mu_0 N^2 \pi R}{2}$$

Answer

$$\frac{\mu_0 N \pi R}{2}$$

Electromagnetic Induction

A circular coil of radius '$R' has 'N' turns of a wire. The coefficient of self induction of the coil will be ( \mu_0=$ permeability of free space)

Choose the correct answer: