A magnetic needle suspended parallel to a magnetic field requires $square root of 3$ J of work to turn it through 60^o. The torque needed to maintain the needle in this position will be

$2 square root of 3 space straight J$

3 J

$square root of 3 space J$
$3 over 2 space straight J$

Solution

3 J

In this case, work done
$straight W space equals space MB space left parenthesis cosθ subscript 1 minus Cosθ subscript 2 right parenthesis equals space MB space left parenthesis cos space 0 to the power of straight o minus cos space 60 to the power of straight o right parenthesis equals space MB space open parentheses 1 minus 1 half close parentheses space equals space open parentheses MB over 2 close parentheses MB space equals space 2 square root of 3 space straight J space space space space space space space left parenthesis therefore space given space straight W space equals space square root of 3 space straight J right parenthesis straight zeta space equals space MB space sin space 60 to the power of straight o space equals space left parenthesis 2 square root of 3 right parenthesis open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses space straight J space equals space 3 space straight J$