Answer :

Potential energy of a loop in a uniform magnetic field is given by


U=-MBcosθ where,


M is magnetic moment of the current carrying loop,


B is magnetic field in which the loop is kept,


θ is the angle between M and B.


So, the change in potential energy of the loop will give the work done by the magnetic field as initial and final kinetic energy is 0. Since, there is no change in angle between M bad B when we are rotating the loop along the axis perpendicular to its plane so, there will be no change in the potential energy of the loop. And, W= Uf – Ui, where Ui is initial potential energy of the loop and Uf is final energy of the loop. So, W=0.


Hence, our answer is option (d).

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

A bar magnet of mPhysics - Board Papers

Consider a circulPhysics - Exemplar

(a) Derive the exPhysics - Board Papers

A circular currenPhysics - Exemplar

A rectangular conPhysics - Exemplar

A uniform conductPhysics - Exemplar

A uniform magnetiNCERT - Physics Part-I

A square coil of NCERT - Physics Part-I

A circular coil oNCERT - Physics Part-I