Question

Find the magnitude of magnetic field $B$ at the centre of a bar magnet having pole strength $3.6\text{Am}$, magnetic length $12\text{cm}$ and cross-sectional area $0.90{\text{cm}}^2$.

• A. $4.0\times {10}^{-2}\text{T}$
• B. $5.0\times {10}^{-2}\text{T}$
• C. $6.0\times {10}^{-2}\text{T}$
• D. $7.0\times {10}^{-2}\text{T}$

Answer 1

The correct option is B $5.0\times {10}^{-2}\text{T}$

Given:
Pole strength of magnet,$m=3.6\text{Am}$,

Magnetic length, $2l=12\text{cm}$

Cross-sectional area, $A=0.9{\text{cm}}^2$

The magnetic field at the centre of the magnet is,

$\overrightarrow{B}={\mu }_0(\overrightarrow{I}+\overrightarrow{H})$

Here,

The magnetization of the given bar magnet is,

$I=\frac{M}{V}=\frac{m\left(2l\right)}{A\left(2l\right)}=\frac{m}{A}$

$I=\frac{3.6}{0.90\times {10}^{-4}}=4\times {10}^4\text{A/m}\left[\text{Towards the North Pole}\right]$


Magnetic intensity due to north pole at the centre is,

$H_N=\frac{1}{4\pi }\frac{m}{l^2}=\frac{1}{4\pi }\frac{3.6}{(6\times {10}^{-2}{)}^2}=\frac{1000}{4\pi }\text{A/m}$

Similarly, magnetic intensity due to south pole at the centre is,

$H_S=\frac{1000}{4\pi }\text{A/m}$


Resultant magnet intensity,

$H=H_N+H_S=\frac{2000}{4\pi }=159.2\text{A/m}\left[\text{Towards the South Pole}\right]$


Using the equation $\left(1\right)$, the net magnitude of magnetic field at the centre is,

$B={\mu }_0(I-H)$

$B=(4\pi \times {10}^{-7})(4\times {10}^4-159.2)$

$=5.0\times {10}^{-2}\text{T}\left[\text{Towards North Pole}\right]$

Hence, option $\left(B\right)$ is correct.