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Q.A bar magnet is initially at right angles to a uniform magnetic field. The magnet is rotated till the torque acting on it becomes one-half of its initial value. The angle through which the bar magnet is rotated is: (A) 30∘30^\circ (B) 45∘45^\circ (C) 60∘60^\circ (D) 75∘75^\circ

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The torque on a magnetic dipole is τ=mBsin⁡θ\tau = mB\sin\theta, maximum when perpendicular (θ=90∘\theta = 90^\circ). When torque drops to half its initial value, sin⁡θ=12\sin\theta = \frac{1}{2}, giving θ=30∘\theta = 30^\circ — so the magnet rotates through 60∘60^\circ.

Understanding torque on a magnetic dipole

When a bar magnet (magnetic dipole of moment mm) sits in a uniform magnetic field BB, it experiences a torque that tries to align it with the field. The magnitude of this torque depends on how misaligned the dipole is:

τ=mBsin⁡θ\tau = mB\sin\theta

where θ\theta is the angle between the magnetic moment vector and the field direction.

The torque is maximum when the dipole is perpendicular to the field (θ=90∘\theta = 90^\circ, so sin⁡θ=1\sin\theta = 1), and zero when aligned (θ=0∘\theta = 0^\circ). This makes physical sense: the restoring couple is strongest when the magnet is sideways to the field lines.

τ=mBsin⁡θ\tau = mB\sin\theta

Step-by-step solution

1. Identify the initial configuration

The magnet starts "at right angles to a uniform magnetic field," meaning the magnetic moment makes an angle θi=90∘\theta_i = 90^\circ with the field. The initial torque is:

τi=mBsin⁡90∘=mB\tau_i = mB\sin 90^\circ = mB

2. Set up the condition after rotation

After rotating the magnet, the torque becomes half the initial value:

τf=τi2=mB2\tau_f = \frac{\tau_i}{2} = \frac{mB}{2}

Let the new angle between the magnetic moment and field be θf\theta_f. Then:

mBsin⁡θf=mB2mB\sin\theta_f = \frac{mB}{2}

3. Solve for the final angle

Dividing both sides by mBmB:

sin⁡θf=12\sin\theta_f = \frac{1}{2}

This gives: …

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