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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
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Verified Answer
The correct answer is:
$45^{\circ}$
As we know, torque in a magnetic field,
$$
\begin{aligned}
\tau & =M B \sin \theta \\
\theta & =\theta_l,
\end{aligned}
$$
Similarly, if $\theta=\theta_2$ then
$$
\begin{aligned}
\tau_2=M B \sin \theta_2 & =M B \sin 2 \theta_1 \\
\text { Given } \quad \therefore \quad \tau_2 & =\tau_1+\tau_1 \times \frac{41.4}{100} \\
& =1.414 \tau_1=\sqrt{2} \tau_1
\end{aligned} \quad\left(\because \text { Given, } \theta_2=2 \theta_1\right)
$$
From Eqs. (i) and (ii), we get
$$
\frac{1}{\sqrt{2}}=\frac{\sin \theta_1}{\sin 2 \theta_1}
$$
As we know that $\sin 2 \theta=2 \sin \theta \cos \theta$ Hence, $2 \sin \theta_1 \cos \theta_1=\sqrt{2} \sin \theta_1$
$$
2 \cos \theta=\sqrt{2} \Rightarrow \cos \theta=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ}
$$
Hence, the correct option is (d).
$$
\begin{aligned}
\tau & =M B \sin \theta \\
\theta & =\theta_l,
\end{aligned}
$$
Similarly, if $\theta=\theta_2$ then
$$
\begin{aligned}
\tau_2=M B \sin \theta_2 & =M B \sin 2 \theta_1 \\
\text { Given } \quad \therefore \quad \tau_2 & =\tau_1+\tau_1 \times \frac{41.4}{100} \\
& =1.414 \tau_1=\sqrt{2} \tau_1
\end{aligned} \quad\left(\because \text { Given, } \theta_2=2 \theta_1\right)
$$
From Eqs. (i) and (ii), we get
$$
\frac{1}{\sqrt{2}}=\frac{\sin \theta_1}{\sin 2 \theta_1}
$$
As we know that $\sin 2 \theta=2 \sin \theta \cos \theta$ Hence, $2 \sin \theta_1 \cos \theta_1=\sqrt{2} \sin \theta_1$
$$
2 \cos \theta=\sqrt{2} \Rightarrow \cos \theta=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ}
$$
Hence, the correct option is (d).
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