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)

As we know, torque in a magnetic field,
$\tau =MB\sin \theta$
if $\theta ={\theta }_1,$
then ${\tau }_1=MB\sin {\theta }_1\dots$(i)
Similarly, if $\theta ={\theta }_2$ then
${\tau }_2=MB\sin {\theta }_2=MB\sin 2{\theta }_1$
$(\because$ Given, ${\theta }_2=2{\theta }_1)$
Given $\therefore {\tau }_2={\tau }_1+{\tau }_1\times \frac{41.4}{100}$
$=1.414{\tau }_1=\sqrt{2}{\tau }_1$
$\Rightarrow \sqrt{2}{\tau }_1=MB\sin 2{\theta }_1\dots$(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, $2\sin {\theta }_1\cos {\theta }_1=\sqrt{2}\sin {\theta }_1$
$2\cos \theta =\sqrt{2}$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}$
$\Rightarrow \theta =45^{\circ }$