Q. A magnetic needle lying parallel to a magnetic field requires $W$ unit of work to turn it through $60^\circ $ . What is the torque needed to maintain the needle in this position ?
NTA AbhyasNTA Abhyas 2020
Solution:
The work done in rotating the coil will be;
$W=MB\left(C o s \left(\theta \right)_{1} - C o s \left(\theta \right)_{2}\right)$
$=MB\left(C o s 0 ^\circ - C o s 60 ^\circ \right)$
$=MB\left(1 - \frac{1}{2}\right)=\frac{M B}{2}$
The torque needed to maintain the coil in this position will be;
$\tau=MBSin\theta =MBSin60^\circ =\frac{\sqrt{3}}{2}MB$
$=\frac{\sqrt{3}}{2}\left(\right.2W\left.\right)\left[\because \frac{M B}{2} = W\right]$
$\Rightarrow \tau=\sqrt{3}W$
$W=MB\left(C o s \left(\theta \right)_{1} - C o s \left(\theta \right)_{2}\right)$
$=MB\left(C o s 0 ^\circ - C o s 60 ^\circ \right)$
$=MB\left(1 - \frac{1}{2}\right)=\frac{M B}{2}$
The torque needed to maintain the coil in this position will be;
$\tau=MBSin\theta =MBSin60^\circ =\frac{\sqrt{3}}{2}MB$
$=\frac{\sqrt{3}}{2}\left(\right.2W\left.\right)\left[\because \frac{M B}{2} = W\right]$
$\Rightarrow \tau=\sqrt{3}W$