Q .     A magnetic needle lying parallel to the
magnetic field requires `W` units of work to
turn it through an angle `45°`. The torque
required to maintain the needle in this
position will be

VIT 2015
A

` sqrt2 W`

B

`1/(sqrt(3)W)`

C

`(sqrt(2)- 1)W`

D

`W/((sqrt(2) -1)`

HINT

Work done by a magnet to tum from angle `theta_(1)` to `theta_(2)` is `W = MB(costheta_(1) - costheta_(2) )` ` tau =MBsin45° `

Today's Score

0
Success Rate: NAN %

Practice & Test

Sr. No.SubjectTagCorrectRecorded Answer