A magnetic needle is in the most equilibrium state in a uniform magnetic field requires "W" units of work to turn through 60°. The work done to turn the magnetic needle from the position of most stable equilibrium to the most unstable equilibrium position is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4W
The potential energy is: $U(θ)=-mB\cos θ$
The work done in rotating the needle from $θ_1$ to $θ_2$,
$W=U(θ_1)-U(θ_2)$
$=U(0°)-U(60°)$
$=-mB(1)+mB(\frac{1}{2})=\frac{mB}{2}=W$
$W_{unstable}=U(0°)-U(180°)$
$=(-mB\cos 0°)-(-mB(-1))$
$=2mB$
$∴W_{unstable}=2mB=2×2W=4W$ $[mB=2W]$