Question

The work required to rotate a magnetic needle by 60${}^0$ from equilibrium position in a uniform magnetic field is W. The torque required to hold it in that position is

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

Answer 1

Answer: (D)

$\sqrt{3}W$

${\theta }_1=0$
${\theta }_2=60$
$U_1=-mBcos0$
$=-mB$
$U_2=-mBcos{60}^0$
$=-\frac{mB}{z}$
$w=U_2-U_1$
$=\frac{mB}{z}$
$z=mBsin\theta$
$=mBsin60$
$=\frac{mB}{z}\sqrt{3}$
$z=w\sqrt{3}$