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Question

(D) The torque $	au$ experienced by a bar magnet in a uniform magnetic field $B$ is given by $	au = MB \sin 	heta$,where $M$ is the magnetic dipole

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(D) The torque $	au$ experienced by a bar magnet in a uniform magnetic field $B$ is given by $	au = MB \sin 	heta$,where $M$ is the magnetic dipole moment and $	heta$ is the angle between the magnet and the field.Let the initial angle be $	heta_1$. Then the initial torque is $	au_1 = MB \sin 	heta_1$ $(i)$.If the angle is doubled,the new angle is $	heta_2 = 2	heta_1$. The new torque is $	au_2 = MB \sin 	heta_2 = MB \sin 2	heta_1$ (ii).Given that the torque increases by $41.4 \%$,we have $	au_2 = 	au_1 + 0.414 	au_1 = 1.414 	au_1$.Since $\sqrt{2} \approx 1.414$,we can write $	au_2 = \sqrt{2} 	au_1$.Substituting the expressions for $	au_1$ and $	au_2$ from $(i)$ and (ii):$MB \sin 2	heta_1 = \sqrt{2} MB \sin 	heta_1$Using the trigonometric identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$:$2 \sin 	heta_1 \cos 	heta_1 = \sqrt{2} \sin 	heta_1$Assuming $\sin 	heta_1 
eq 0$,we divide both sides by $\sin 	heta_1$:$2 \cos 	heta_1 = \sqrt{2}$$\cos 	heta_1 = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$Therefore,$	heta_1 = 45^{\circ}$.Thus,the correct option is $D$.

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