1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A bar magnet of magnetic moment 6 J/T is aligned at 600 with a uniform external magnetic field of 0.44 T. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field, (ii) opposite to the magnetic field, and (b) the torque on the magnet in the final orientation in case (ii).

Open in App
Solution

## Given, magnetic momentum, m = 6 JT−1 External magnetic field, B = 0.44 T θ1=600⇒cos θ1=cos 600=12. (a) Work done in turning the magnet normal to the field, W=−mB(cos θ2−cos θ1) (i) Here, θ2=900 ∴ W=+mB cosθ1 = 6×0.44×12=1.32 J (ii) Here θ2=1800 ∴ W=−mB(cos θ2−cos θ1) W=−6×0.44(−1−12) = 3.96 J (b) Torque on magnet when moment is aligned opposite to the field, τ=mB sin θ putting the values, = 6×0.44×sin 180∘ = 0(∵sin 180∘=0)

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Torque on a Magnetic Dipole
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program