First slide

Magnetic dipole in a uniform magnetic field

Question

If a magnet of length 10 cm and pole strength 40 Am is placed at an angle of $45^{o}$ in an uniform induction field of intensity $2 x 10^{- 4} T$ , the couple acting on it is

Easy

A

$0.5656 x 10^{- 4} Nm$
B

$0.5656 x 10^{- 3} Nm$
C

$0.656 x 10^{- 4} Nm$
D

$0.656 x 10^{- 5} Nm$

Solution

$τ = MB \sin θ = (mL) B \sin θ$
$= (40 x 10 x 10^{- 2}) x 2 x 10^{- 4} \sin 45^{o}$
$= 0.565 x 10^{- 3} N m$

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