Question

A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through ${60}^o$. The torque needed to maintain the needle in this position will be

• A. $\sqrt{3}W$
• B. $W$
• C. $\frac{\sqrt{3}W}{2}$
• D. $2W$

Answer 1

Answer: (A)

$\sqrt{3}W$

${\theta }_1=0^0$
${\theta }_2={60}^0$
$U_1=mBcoso$
$=-mB$
$U_2=-mBcos6$
$=\frac{-mB}{z}$
$w=U_2-U_1$
$=\frac{mB}{z}$
$z=mBsin60$
$z=mB\frac{\sqrt{3}}{2}$
$z=w\sqrt{3}$