First slide

Magnetic dipole in a uniform magnetic field

Question

A bar magnet of length 10 cm and having the pole strength equal to 10^-3 Wb is kept in a magnetic field having magnetic induction equal to 4 $π$ x 10^-3 T. It makes an angle 30^o with the direction of magnetic induction. The value of the torque acting on the magnet is $μ_{0} = 4 π \times 10^{- 7} {WbA}^{- 1} m$

Moderate

A

$2 π \times 10^{- 7} Nm$
B

$2 π \times 10^{- 5} Nm$
C

0.5 Nm
D

$0.5 \times 10^{- 2} Nm$

Solution

Pole strength = $10^{- 3} Wb = \frac{10^{- 3}}{μ_{0}} Am$
Magnetic moment, M = m(2l)
$∴ M = \frac{10^{- 3}}{μ_{0}} \times 0 \cdot 10 = \frac{10^{- 4}}{μ_{0}} Now τ = MBsin θ \begin{array}{l} = \frac{10^{- 4}}{μ_{0}} \times (4 π \times 10^{- 3}) \times \sin 30^{\circ} \\ = \frac{10^{- 4}}{(4 π \times 10^{- 7})} \times (4 π \times 10^{- 3}) \times \frac{1}{2} = \frac{1}{2} = 0 \cdot 5 Nm \end{array}$

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