If a bar magnet of moment is suspended in a uniform magnetic field $¯ ¯¯ ¯ B$ it is given an angular deflection, w.r.t equilibrium position. Then the restoring torque on the magnet is

The work required to rotate a magnetic needle by 60 $^{0}$ from equilibrium position in a uniform magnetic field is W. The torque required to hold it in that position is

Join BYJU'S Learning Program

A bar magnet has a magnetic moment of 200 Am2. The magnet is suspended in a magnetic field of 0.30 NA−1m−1. The torque required to rotate the magnet from its equilibrium position through an angle of 30o will be:

The correct option is D 30 NmGiven M=200 Am2;B=0.30 NA−1M−1and θ=30oWe know that the torque→τ=→M×→B⇒|τ|=MBsinθ=200×0.3×12=100×0.3=30 Nm

A bar magnet has a magnetic moment of $200$ $A m^{2}$ . The magnet is suspended in a magnetic field of $0.30 N A^{- 1} m^{- 1}$ . The torque required to rotate the magnet from its equilibrium position through an angle of $30^{o}$ will be:

The correct option is D $30 N m$
Given $M = 200 A m^{2}; B = 0.30 N A^{- 1} M^{- 1}$
and $θ = 30^{o}$
We know that the torque
$\to τ = \to M \times \to B$
$\Rightarrow | τ | = M B sin θ = 200 \times 0.3 \times \frac{1}{2}$
$= 100 \times 0.3 = 30 N m$