Question

A bar magnet of magnetic moment 6 J/T is aligned at 60° 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). 

Answer 1

(a) Formula and Calculation of work done in the two cases

(b) Calculation of torque in case (ii)

(a) Work done =  m_B(cosθ₁− cosθ₂

(i)  θ₁ = 60°, θ₂ = 90°

[Also accept calculations done through changes in potential energy.]

(b)  

[If the student straight away writes that the torque is zero since magnetic moment and magnetic field are anti parallel in this orientation, award full]

Detailed Answer :

(a) (i) θ₁ = 60° , θ₂ = 90° since magnet is placed perpendicular to magnetic field. So, work done in rotating the magnet from  θ₁ to θ₂ is

(ii) Work done in aligning the magnet opposite to magnetic field. i.e  θ₂ = 180° , θ₁ = 60°

(b) The Torque on magnet aligned at angle θ₂ is given by t =

Torque in case a (ii) i.e. at θ₂ = 180 position is zero.