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A bar magnet of magnetic moment ⃗M is placed in a magnetic field of induction ⃗B. The torque exerted on it is

Asked in GUJCET 2007 · Dipole in a uniform field

Answer: (2) ⃗M×⃗B

Step-by-step solution

Idea: a uniform field pushes the north pole one way and the south pole the other with equal and opposite forces. The net force is zero, but the pair of forces twists the magnet.

That couple is the torque τ⃗=⃗M×⃗B, of magnitude τ=MB sin θ.

The cross product also fixes the direction: τ⃗ is perpendicular to both ⃗M and ⃗B, and it turns the magnet towards the field.

Check the two limits: at θ=0 (aligned) τ=0, and at θ=90° τ=MB, the largest value.

The other three options are dot products, which are scalars; -⃗M·⃗B is the potential energy.

So the torque exerted on the magnet is ⃗M×⃗B.

Why the other options are wrong

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