Question

A circular coil of radius $R$ and a current $I$, which can rotate about a fixed axis passing through its diameter is initially placed such that its plane lies along magnetic field $B$. Kinetic energy of loop when it rotate through an angle ${90}^0$ is:(Assume that $I$ remain constant)

• A. $\pi R^{2}BI$
• B. $\frac{\pi R^{2}BI}{2}$
• C. $2\pi R^{2}BI$
• D. $\frac{3}{2}\pi R^{2}I$

Answer 1

The correct option is A $\pi R^{2}BI$

Here, change in kinetic energy is equal to the potential energy difference between the initial and final positions. So, when the plane is parallel to the magnetic field direction, the angle between the area vector and the magnetic field vector is 90 So, the potential energy is Zero and when the coil is rotated by 90 the potential energy is,

$U=-MBcos\theta$ $=-MB$ ($\therefore$ $cos90=1$)

$So,$ $the$ $kinetic$ $energy$ $is$

$K=$ $△U$ $=MB$ $=IAB$ $=\pi R^{2}IB$