Question

The work required to be done in rotating s magnetic needle placed in uniform magnetic field by ${60}^0$ from.its equilibrium position is W, the torque required so that the needle remains in this position is:-

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

Answer 1

The correct option is B $W\sqrt{3}$

Initially

angle between needle and field

$=0^{\circ }$

finally it changes to ${60}^{\circ }$

$W=△U=M\times B\left(cos\right(0)-cos60)$

$=\frac{MB}{2}$

$M.B=2W$____ (1)

Now

Torque acting on needle

$I=M\times B$

$I=MBsin{60}^{\circ }$

$I=2W\times \frac{\sqrt{3}}{2}$

$I=W\sqrt{3}$