First slide

Magnetic dipole in a uniform magnetic field

Question

A bar magnet when placed at an angle of 30^o to the direction of magnetic field induction of 5 x 10^-4 T, experiences a moment of couple 25 x 10^-6 Nm. If the length of the magnet is 5 cm, its pole strength is

Easy

A

2 x 10^-2 Am
B

5 x 10^-2 Am
C

2 A m
D

5 A m

Solution

We know that, $τ = MBsin θ$
$∴ M = \frac{τ}{Bsin θ} = \frac{(25 \times 10^{- 6})}{(5 \times 10^{- 4}) \times (1 / 2)} = 10 \times 10^{- 2} {Am}^{2} Now m = \frac{M}{L} = \frac{10 \times 10^{- 2}}{0 \cdot 05} = 2 Am$

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