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A circular coil of 10 turns and radius $10 \mathrm{~cm}$ is placed in a uniform magnetic field of $0.1 \mathrm{~T}$ normal to the plane of the coil. If the current in the coil is $5 \mathrm{~A}$, then the magnitude of the torque on the coil is
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The correct answer is:
Zero
Magnitude of torque on the coil is given as
$\tau=|\mathbf{m} \times \mathbf{B}| \Rightarrow \tau=N I A B \sin \theta$
where, $\theta$ is the angle between axis of loop (normal to the loop) and direction of magnetic field.
Here, $\theta=0 \quad$ So, $\quad \tau=0$
$\tau=|\mathbf{m} \times \mathbf{B}| \Rightarrow \tau=N I A B \sin \theta$
where, $\theta$ is the angle between axis of loop (normal to the loop) and direction of magnetic field.
Here, $\theta=0 \quad$ So, $\quad \tau=0$
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