A short bar magnet placed in a uniform magnet | vedclass

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

Vedclass · Accepted Answer

increases by $41.4 \%$

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(D) The torque $	au$ on a bar magnet in a uniform magnetic field $B$ is given by the formula $	au = MB \sin 	heta$, where $M$ is the magnetic dipole moment and $	heta$ is the angle between the magnetic moment and the magnetic field.Initial torque $	au_1 = MB \sin 30^{\circ} = MB 	imes 0.5$.Final torque $	au_2 = MB \sin 45^{\circ} = MB 	imes \frac{1}{\sqrt{2}} \approx MB 	imes 0.707$.The ratio of the torques is $\frac{	au_2}{	au_1} = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2} \approx 1.414$.Therefore, $	au_2 = 1.414 	au_1$.The percentage increase in torque is given by $\frac{	au_2 - 	au_1}{	au_1} 	imes 100 = (1.414 - 1) 	imes 100 = 41.4 \%$.Thus, the torque increases by $41.4 \%$.

$A$ short bar magnet placed in a uniform magnetic field making an angle with the field experiences a torque. If the angle made by the magnet with the field is changed from $30^{\circ}$ to $45^{\circ}$, the torque of the magnet

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