Question

The magnetic moment of a short bar magnet is $2A{m}^2$. The magnetic field at a point $1m$ away it in a direction making an angle ${60}^o$ with its magnetic moment is /;

• A. $\sqrt{5}\times {10}^{-7}T$
• B. ${10}^{-7}T$
• C. $\sqrt{7}\times {10}^{-7}T$
• D. $7\times {10}^{-7}T$

Answer 1

Answer: (C)

$\sqrt{7}\times {10}^{-7}T$

Magnetic field due to a bar magnet of magnetic dipole moment $M$ at a

distance from it in a direction $\theta$ with its magnetic moment is

$B=\frac{{\mu }_0}{4\pi }.\frac{M}{r^3}\sqrt{(3{\cos }^2\theta +1)}$

where,

$\frac{{\mu }_0}{4\pi }={10}^{-7}Tm/A$

Given,

$M=2A{m}^2$

$r=1m$

$\theta ={60}^0$

$\Rightarrow B=\left({10}^{-7}Tm/A\right)\times \frac{2A{m}^2}{(1m{)}^3}\times \sqrt{(\frac{3}{4}+1)}$

$=\sqrt{7}\times {10}^{-7}T$