Question

A bar magnet placed in a uniform magnetic field making an angle $\theta$ with the field experiences a torque. If the angle made by the magnet with the field is doubled, the torque experienced by the magnet increases by $41.4\%$. The initial angle made by the magnet with the magnetic field is

• A. $60^{\circ}$
• B. $30^{\circ}$
• C. $90^{\circ}$
• D. $45^{\circ}$

Answer 1

Answer: (D)

$45^{\circ}$

Answer 2

As we know, torque in a magnetic field,
$\tau =M B\, \sin \,\theta$
if $\theta=\theta_{1},$
then $\tau_{1}=M B\, \sin\, \theta_{1} \dots$(i)
Similarly, if $\theta=\theta_{2}$ then
$\tau_{2}=M B\, \sin\, \theta_{2}=M B\, \sin \,2 \theta_{1}$
$\left(\because\right.$ Given, $\left.\theta_{2}=2 \theta_{1}\right)$
Given $\therefore \tau_{2}=\tau_{1}+\tau_{1} \times \frac{41.4}{100}$
$=1.414 \tau_{1}=\sqrt{2} \tau_{1}$
$\Rightarrow \sqrt{2} \tau_{1}=M B\, \sin \,2 \theta_{1} \dots$(ii)
From Eqs. (i) and (ii), we get
$\frac{1}{\sqrt{2}}=\frac{\sin\, \theta_{1}}{\sin \,2 \theta_{1}}$
$\Rightarrow \sin \,2 \theta_{1}=\sqrt{2} \sin\, \theta_{1} \dots$(iii)
As we know that $\sin \,2 \theta=2 \,\sin \,\theta \,\cos\, \theta$
Hence, $2 \,\sin \,\theta_{1} \,\cos \,\theta_{1}=\sqrt{2} \sin \,\theta_{1}$
$2 \,\cos \,\theta=\sqrt{2}$
$\Rightarrow \cos\, \theta=\frac{1}{\sqrt{2}}$
$\Rightarrow \theta=45^{\circ}$