Question

Question 5.3: A short bar magnet placed with its axis at 30º with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10 −2 J. What is the magnitude of magnetic moment of the magnet?

Class 12 - Physics - Magnetism And Matter Magnetism And Matter Page-200

Answer 1

we know, $\vec{\tau}=\vec{M}\times\vec{B}$
for magnitude, $\tau=MBsin\theta$
here, $\tau$ is the torque due to bar magnet, M is the magnetic moment of bar magnet and B is the magnetic field .

Here, $\tau=4.5\times10^{-2}J$
B = 0.25T, $\theta=30^{\circ}$
so, 4.5 × 10^-2 = M × 0.25 × sin30°
=> 4.5 × 10^-2 = M × 1/4 × 1/2
=> 4.5 × 10^-2 × 8 = M
=> M = 0.36 J/T
hence, magnetic moment = 0.36J/T