A short bar magnet placed with its axis at 30° 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?

Question

Accepted Answer

(1) 0.36 J T⁻¹. Given: θ=30°, B=0.25 T, τ=4.5×10⁻² J (a joule here is a newton metre). Idea: the torque on a dipole in a uniform field is τ=mB sin θ, so m=τ/(B sin θ). sin 30°=0.5, so B sin θ=0.25×0.5=0.125 T. m=(4.5×10⁻²)/(0.125)=0.36 J T⁻¹. Check by going back: 0.36×0.125=4.5×10⁻² J, the given torque. So the magnetic moment of the magnet is 0.36 J T⁻¹.

A short bar magnet placed with its axis at 30° 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?

Step-by-step solution

Why the other options are wrong