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If the solenoid in Exercise $5.5$ is free to turn about the vertical direction and a uniform horizontal magnetic field of $0.25 \mathrm{~T}$ is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of $30^{\circ}$ with the direction of applied field?
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Verified Answer
Given: Horizontal magnetic field $\mathrm{B}=0.25 \mathrm{~T}$, magnetic moment (from Q. 5.5) $\mathrm{M}=0.6 \mathrm{Am}^2$ angle $\theta=30^{\circ}$
To find: Torque $\tau$
Formula used: $\tau=\mathrm{MB} \sin \theta$
$$
\tau=0.6 \times 0.25 \times \sin 30^{\circ}=0.15 \times \frac{1}{2}=0.075 \mathrm{Nm}
$$
To find: Torque $\tau$
Formula used: $\tau=\mathrm{MB} \sin \theta$
$$
\tau=0.6 \times 0.25 \times \sin 30^{\circ}=0.15 \times \frac{1}{2}=0.075 \mathrm{Nm}
$$
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