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-30sqrt - 3 - Nm60 Nm60sqrt - 3 - Nm30 Nm

Question

Toppr Admin · Accepted Answer

Given M-200-xA0-Am2-B-0-30-xA0-NA-x2212-1M-x2212-1and -x3B8-30oWe know that the torque-x2192-x3C4-x2192-M-xD7-x2192-B-x21D2-x3C4-MBsin-x3B8-200-xD7-0-3-xD7-12-100-xD7-0-3-30-xA0-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:
$30 \sqrt{3} N m$
$60 N m$
$60 \sqrt{3} N m$
$30 N m$

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

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:30√3 Nm60 Nm60√3 Nm30 Nm

Given 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:
$30 \sqrt{3} N m$
$60 N m$
$60 \sqrt{3} N m$
$30 N m$

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