A bar magnet is held perpendicular to a uniform field. If the couple acting on the magnet is to be halved by rotating it from above position, the angle by which it is to be rotated is:

Question

A bar magnet is held perpendicular to a uniform field. If the couple acting on the magnet is to be halved by rotating it from above position, the angle by which it is to be rotated is: (A)  30°  (B)  60°  (C)  40°  (D)  90°

Tardigrade Admin · Accepted Answer

Since, couple acting on the magnet is halved. Hence, τ =MB sin θ [Since, magnet is same here. Mare constant B is same] (1/2)=MB sin θ sin θ =(1/2)= sin 30o θ =30o θ is angle between magnet and direction of field. Hence, the magnet is to be rotated by =90o-30o=60o

Q. A bar magnet is held perpendicular to a uniform field. If the couple acting on the magnet is to be halved by rotating it from above position, the angle by which it is to be rotated is:

$30^{\circ}$

$60^{\circ}$

$40^{\circ}$

$90^{\circ}$

Q. A bar magnet is held perpendicular to a uniform field. If the couple acting on the magnet is to be halved by rotating it from above position, the angle by which it is to be rotated is:

30∘

60∘

40∘

90∘

Since, couple acting on the magnet is halved. Hence, τ=MBsinθ [Since, magnet is same here. Mare constant B is same] 21​=MBsinθ sinθ=21​=sin30o θ=30o θ is angle between magnet and direction of field. Hence, the magnet is to be rotated by =90o−30o=60o

$30^{\circ}$

$60^{\circ}$

$40^{\circ}$

$90^{\circ}$