First slide

Magnetic dipole in a uniform magnetic field

Question

When a bar magnet is placed at 90^o to a uniform magnetic field, it is acted upon by a couple which is maximum. For the couple to be half of the value, the magnet should be inclined to the magnetic field at an angle of

Easy

A

45^o
B

30^o
C

15^o
D

0^o

Solution

$τ = MBsin θ or τ \propto \sin θ (τ_{1} / τ_{2}) = (\sin θ_{1}) / \sin θ_{2} or \frac{τ \times 2}{τ_{1}} = \frac{1}{\sin θ_{2}} or \sin θ_{2} = 1 / 2 ∴ θ_{2} = 30^{\circ}$

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