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Bar Magnet As An Equivalent Solenoid - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Bar magnet as an equivalent solenoid is considered one of the most asked concept.
25 Questions around this concept.

Solve by difficulty

Two coaxial solenoids of different radii carry current I in the same direction. Let $\underset{F_{1} }{\rightarrow}$ be the magnetic force on the inner solenoid due to the outer one and $\underset{F_{2} }{\rightarrow}$ be the magnetic force on the outer solenoid due to the inner one. Then :

$\underset{F_{1} }{\rightarrow}$ = $\underset{F_{2} }{\rightarrow}$ = 0

$\underset{F_{1} }{\rightarrow}$ is radially inwards and $\underset{F_{2} }{\rightarrow}$ radially outwards

$\underset{F_{1} }{\rightarrow}$ is radially inwards and $\underset{F_{2} }{\rightarrow}$ = 0

$\underset{F_{1} }{\rightarrow}$ is radially outwards $\underset{F_{2} }{\rightarrow}$ = 0

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Concepts Covered - 1

Bar magnet as an equivalent solenoid

Bar Magnet-

A bar magnet consists of two equal and opposite magnetic pole separated by a small distance.

Pole strength (m)-

The strength of a magnetic pole to attract magnetic materials towards itself is known as pole strength.

It is a scalar quantity and it is represented by +m and -m.

It depends on the nature of the material of magnet and area of cross-section i.e, independent from the length.

Magnetic dipole moment ( $\vec{M}$ )- It represents the strength of the magnet. Mathematically it is defined as
the product of the strength of either pole and effective length.

i.e for the below figure $\vec{M}=mL=m(2l)$

Fig 5

It is a vector quantity directed from south to north.

This is analogous to electrical dipole moment which was given by $\vec{P}=qL$

And using this analogy we can calculate

The magnetic field on Axial Position of a bar magnet-

Axial Position

$For \ \ r >> a \ \ \ \Rightarrow B_{axial}= \frac{\mu _{o}\:2M}{4\pi \:r^{3}}$

Magnetic Field at the equatorial position of a magnet-

Equatorial position

$B_{e}= \frac{\mu _{o}}{4\pi }\frac{M}{\left ( r^{2}+a^{2} \right )^{\frac{3}{2}}}$

And for $For \ \ r >> a \ \ \ \Rightarrow B_{e}= \frac{\mu _{o}\:M}{4\pi \:r^{3}}$

Magnetic Field at any general point due to bar magnet

$B_{g}= \frac{\mu _{o}}{4\mu }\frac{M}{r^{3}} \sqrt{3cos^{2}\Theta +1}$

Solenoid-

The solenoid is defined as a cylindrical coil of many tightly wound turns of insulated wire with generally diameter of the coil smaller than its length.

Bar magnet as an equivalent solenoid-

By calculating the axial field of a finite solenoid carrying current and equating it with the magnetic field of bar magnet we can demonstrate a Bar magnet as an equivalent solenoid.

For the above figure

Let n = number of turns per unit length $\frac{N}{L}$

whrere, N = total number of turns,

$L=2l$ = length of the solenoid

We will take an elemental circular current-carrying loop of thickness dx and radius R at a distance x from the center of the solenoid.

So the number of turns per unit length for elemental loop will be $n=\frac{N}{dx}$

And the magnetic field at point P due to an elemental loop is given as $d B=\frac{\mu_{0}(n d x) I R^{2}}{2\left\{(r-x)^{2}+R^{2}\right\}^{3 / 2}}$

for $r >> R \ \ \ and \ \ r >>x$

$d B=\frac{\mu_{0}(n d x) I R^{2}}{2r^3}$

Integrating x from $-l \ \ to \ \ +l$ we get the magnitude of the total field as

$B= \int_{-l}^{l} \frac{\mu _{0} n I R^{2}}{2 r^{3}} d x=\frac{\mu _{0} n I R^{2}}{2 r^{3}} \int_{-l}^{l}dx= \frac{\mu _{0} n I R^{2}}{2 r^{3}} *(2l)\\ Now \ divide \ and \ multyply \ by \ \pi \\ \Rightarrow \vec{B}=\frac{\mu _{0} (n2l) I\pi R^{2}}{2\pi r^{3}}$

Using $N=n(2l)$

we get $\vec{B}=\frac{\mu _{0} N I\pi R^{2}}{2\pi r^{3}}$

Now if we consider above solenoid as a Bar magnet then its dipole moment is given by $\vec{M}=NIA$

Now using $A=\pi R^2$ we can write $\vec{B}=\frac{\mu _{0} N IA}{2\pi r^{3}}=\frac{\mu _{0} \vec{M}}{2\pi r^{3}}=\frac{2\mu _{0} \vec{M}}{4\pi r^{3}}$