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The work done in turning a magnet of mag...

The work done in turning a magnet of magnetic moment 'M' by an angle of `90^(@)` from the meridian is 'n' times the corresponding work done to turn it through an angle of `60^(@)`, where 'n' is given by

A

`1//2`

B

2

C

`1//4`

D

1

Text Solution

Verified by Experts

The correct Answer is:
A
`W_(1)= - MB (cos 90^(@)- cos 0^(@)) = MB` ...(i)
`W_(2)= - MB (cos 60^(@)- cos 0^(@))= (MB)/(2)` ...(ii)
By equation (i) and (ii)
`n= (1)/(2) ( because W_(1)= n W_(2))`
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