Question

When a bar magnet is placed at 90${}^o$ in a uniform magnetic field, then it is acted upon by a maximum couple. For the couple to be half the maximum value, the magnet should be inclined to the magnetic field at an angle of :

• A. ${45}^o$
• B. ${60}^o$
• C. ${75}^o$
• D. ${30}^o$

Answer 1

Answer: (D)

${30}^o$

We know that the torque on a magnetic dipole placed in a uniform magnetic field is $\overrightarrow{M}\times \overrightarrow{B}$

$\Rightarrow \overrightarrow{\tau }$ $=\overrightarrow{M}\times \overrightarrow{B}$

$|\overrightarrow{\tau }|$ $=MB\sin \theta$, where $\theta$ is the angle between them

For, $\theta ={90}^0$, $|\overrightarrow{\tau }|=Maximum=MB$

So, For the couple to be half the maximum value $i.e\frac{1}{2}MB$

$\Rightarrow$ $MB\sin \theta =$$\frac{1}{2}MB$

$\Rightarrow$ $\sin \theta$ $=\frac{1}{2}$

$\Rightarrow$ $\theta ={30}^0$

Therefore, D is correct option.