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

Bar Magnet-

A bar magnet consists of two equal and opposite magnetic poles 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 the magnet and area of cross-section i.e, independent from the length.

 

Magnetic dipole moment - 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  $\vec{M}=m L=m(2 l)$

It is a vector quantity directed from south to north.

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

And using this analogy we can calculate

• The magnetic field on Axial Position of a bar magnet-

        For $r \gg a \Rightarrow B_{\text {axial }}=\frac{\mu_o 2 M}{4 \pi r^3}$

• Magnetic Field at the equatorial position of a magnet-

$$
B_e=\frac{\mu_o}{4 \pi} \frac{M}{\left(r^2+a^2\right)^{\frac{3}{2}}}
$$

And for For $r \gg a \quad \Rightarrow B_e=\frac{\mu_o M}{4 \pi r^3}$
Magnetic Field at any general point due to bar magnet

$$
B_g=\frac{\mu_o}{4 \mu} \frac{M}{r^3} \sqrt{3 \cos ^2 \Theta+1}
$$
 

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 $\mathrm{n}=$ number of turns per unit length $\frac{N}{L}$
whrere, $\mathrm{N}=$ total number of turns,

$$
L=2 l=\text { length of the solenoid }
$$

We will take an elemental circular current-carrying loop of thickness $d x$ and radius $R$ at a distance $x$ from the center of the So the number of turns per unit length for the elemental loop will be $n=\frac{N}{d x}$

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 $r \gg R$ and $r \gg x$

$$
d B=\frac{\mu_0(n d x) I R^2}{2 r^3}
$$
 

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

$$
B=\int_{-l}^l \frac{\mu_0 n I R^2}{2 r^3} d x=\frac{\mu_0 n I R^2}{2 r^3} \int_{-l}^l d x=\frac{\mu_0 n I R^2}{2 r^3} *(2 l)
$$

Now divide and multyply by $\pi$

$$
\Rightarrow \vec{B}=\frac{\mu_0(n 2 l) I \pi R^2}{2 \pi r^3}
$$

Using $N=n(2 l)$
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 $\vec{M}=N I A$

Now using $A=\pi R^2$ we can write

$$
\vec{B}=\frac{\mu_0 N I A}{2 \pi r^3}=\frac{\mu_0 \vec{M}}{2 \pi r^3}=\frac{2 \mu_0 \vec{M}}{4 \pi r^3}
$$

$\vec{B}=\frac{2 \mu_0 \vec{M}}{4 \pi r^3}$ This is equivalent to the magnetic field on the Axial Position of a bar magnet.