Question

A short bar magnet of the magnetic moment $6.4{\text{Am}}^2$ has the magnetic field intensity $B$ due to the magnet at a distance $0.4\text{m}$ from the centre of magnet situated on a line making an angle of ${45}^{\circ }$ with magnetic dipole axis.

• A. $B=2.5\times {10}^{-5}\text{T}$
• B. $B=\sqrt{2.5}\times {10}^{-5}\text{T}$
• C. $B$ will make ${\tan }^{-1}(\frac{1}{2})$ angle with magnetic dipole axis.
• D. $B$ will make $\frac{\pi }{4}+{\tan }^{-1}(\frac{1}{2})$ angle with magnetic dipole axis.

Answer 1

Answer: (B), (D)

$B$ will make $\frac{\pi }{4}+{\tan }^{-1}(\frac{1}{2})$ angle with magnetic dipole axis.

Given:
$M=6.4{\text{Am}}^2;r=0.4\text{m};\theta ={45}^{\circ }$

As we know that magnetic field at point $\text{P}$ due to short magnet is given by

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

$B=\frac{4\pi \times {10}^{-7}}{4\pi }\frac{6.4}{(0.4{)}^3}\sqrt{1+3{\cos }^2{45}^{\circ }}$

$\therefore B=\sqrt{2.5}\times {10}^{-5}\text{T}$

Now using relation of $\theta \&\varphi$,

$\tan \varphi =\frac{\tan \theta }{2}=\frac{\tan {45}^{\circ }}{2}=\frac{1}{2}$

$\therefore \varphi ={\tan }^{-1}(\frac{1}{2})$

So, the magnetic field $B$ will make $(\theta +\varphi )$ angle with magnetic dipole axis.

$\therefore (\theta +\varphi )=\frac{\pi }{4}+{\tan }^{-1}(\frac{1}{2})$

Therefore, options $\left(B\right)$ and $\left(D\right)$ are correct.