Question

When the magnetic dipole is rotated through angle `theta`, work done in rotating dipole is given by -mB cos `theta`. Now, a magnetic needle lying parallel to a magnetic field requires W units of work to tum it through 60°. What will be the value of torque needed to maintain the needle in same position?

पर्याय

• `sqrt3` W

• W

• 0

• `(sqrt3//2)` W

Answer 1

`sqrt3` W

Explanation:

Work done here is equivalent to magnetic potential energy of dipole.

`therefore  "W"= -"mB cos" theta = -"mB" ("cos" theta_2 - "cos" theta _1)`

`= -"mB" ("cos"  60°) - ("cos"  0°)`

`"W" = -"mB" (1/2-1) = "mB"/2`   .... (i)

Now, when `theta` = 60°, torque acting oo dipole should be

`tau = "mB sin" theta = "mB sin"  60° = sqrt3/2 "mB"`

Using (i)

`tau = sqrt3  "W"`