Name: Magnets and Magnetism
Uploaded: 2019-07-19
Duration: 3 min 46 s
Description: In this video, we will learn about the different types of magnets and the magnetic field created by different types of magnets. Learn more.

Question

Derive an expression for the axial field of a solenoid of radius

r

containing

n

turns per unit length and carrying current

i

.

Accepted Answer

Consider a solenoid of length $2 l$ , radius $r$ , carrying current $I$ and number of turns per unit length $n$ . The point $P$ lie outside the solenoid at a distance $a$ .

Consider an element of solenoid of length $d x$ at a distance $x$ from its centre.

This element is a circular current loop having ( $n d x$ ) turns. The magnetic field $d B$ at axial point $P$ due to this current loop is

$d B = \frac{μ_{0} (n d x) I r^{2}}{2 {[r^{2} + (a - x)^{2}]}^{\frac{3}{2}}}$

The total magnetic field due to entire solenoid is

$B = \int d B = \int \frac{μ_{0} (n) I r^{2}}{2 {[r^{2} + (a - x)^{2}]}^{\frac{3}{2}}} d x$

For $a > > r$ and $a > > l$ , we have ${[r^{2} + (a - x)^{2}]}^{\frac{3}{2}} = {[a^{2}]}^{\frac{3}{2}} = a^{3}$

$B = \frac{μ_{0} n I r^{2}}{2 a^{3}} \int_{- l}^{+ l} d x \Rightarrow B = \frac{μ_{0} n I r^{2}}{2 a^{3}} (2 l)$

The magnetic moment of Solenoid, $m = N I A$ , where $N$ is the number of turns, $I$ is solenoid current and $A$ is the cross sectional area of the solenoid coil.

$m = n (2 l) \times I \times π r^{2} \Rightarrow \frac{m}{π} = n (2 l) I \times r^{2}$

$∴ B = \frac{μ_{0}}{2 a^{3}} \times \frac{m}{π} = \frac{μ_{0} m}{2 π a^{3}}$

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