Work done in displacing a magnetic dipole of magnetic moment M (bar magnet) in uniform magnetic field B from an angle θ1 to θ2 is

Question

Work done in displacing a magnetic dipole of magnetic moment M (bar magnet) in uniform magnetic field B from an angle θ1 to θ2 is (A)  W=B(cosθ1-cosθ2)/M  (B)  W=M(cosθ1-cosθ2)/B  (C)  W=M B(cosθ1+cosθ2)  (D)  W=M B(cosθ1-cosθ2)

Tardigrade Admin · Accepted Answer

Work done = Change in potential energy =Uf-Ui =-MB cos θ2-(-MB cos θ1) =MB (cos θ1-cos θ2)

Q. Work done in displacing a magnetic dipole of magnetic moment $M$ (bar magnet) in uniform magnetic field $B$ from an angle $θ_{1}$ to $θ_{2}$ is

$W = B (cos θ_{1} - cos θ_{2}) / M$

$W = M (cos θ_{1} - cos θ_{2}) / B$

$W = M B (cos θ_{1} + cos θ_{2})$

$W = M B (cos θ_{1} - cos θ_{2})$

Work done = Change in potential energy
$= U_{f} - U_{i}$
$= - MB cos θ_{2} - (- MB cos θ_{1})$
$= MB (cos θ_{1} - cos θ_{2})$

Q. Work done in displacing a magnetic dipole of magnetic moment M (bar magnet) in uniform magnetic field B from an angle θ1​ to θ2​ is

W=B(cosθ1​−cosθ2​)/M

W=M(cosθ1​−cosθ2​)/B

W=MB(cosθ1​+cosθ2​)

W=MB(cosθ1​−cosθ2​)