Bar Magnet As An Equivalent Solenoid MCQ - Practice Questions & Answers

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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 $\vec{B}=\frac{\mu _{0} N IA}{2\pi r^{3}}=\frac{\mu _{0} \vec{M}}{2\pi r^{3}}=\frac{2\mu _{0} \vec{M}}{4\pi r^{3}}$