Question 307: 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: