JEE Advanced 2024 Paper 2 · Q04 · Magnetic Field

Question

A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from the same point of the loop. The wire loop has mass $m$ and radius $r$ and it is in a uniform vertical magnetic field $\vec{B}_0$, as shown in the figure. Initially, it hangs vertically downwards, because of acceleration due to gravity $g$, on two conducting supports at $P$ and $Q$. When a current $I$ is passed through the loop, the loop turns about the line $PQ$ by an angle $	heta$ given by

[Figure: A circular loop hangs from two horizontal supports at points P and Q. The two straight tangential ends of the wire rest on the supports; the loop hangs below in the vertical plane. A uniform magnetic field $B_0$ points vertically downwards (parallel to gravity) above the supports. When current $I$ flows, the loop tilts about line PQ by angle $	heta$.]

SolveKar AI · Accepted Answer

The magnetic moment of the loop is $\vec{\mu} = I\vec{A}$, magnitude $|\vec{\mu}| = I\pi r^2$. When the loop hangs vertically, its plane is vertical and $\vec{\mu}$ is horizontal. After tilting by angle $	heta$ about $PQ$, the angle between $\vec{\mu}$ and $\vec{B}_0$ (vertical) is $(90^\circ - 	heta)$.

Magnetic torque about line $PQ$: $	au_{	ext{mag}} = \mu B_0 \sin(90^\circ - 	heta) = I\pi r^2 B_0 \cos	heta$ (acts to rotate loop about $PQ$).

Gravitational restoring torque: the centre of mass of the (uniform) loop is at distance $r$ below $PQ$ along the loop's axis. After tilting, its horizontal displacement from $PQ$ is $r\sin	heta$, so $	au_{	ext{grav}} = mg\,r\sin	heta$.

Equilibrium: $I\pi r^2 B_0 \cos	heta = mg\,r\sin	heta$
$\Rightarrow 	an	heta = \dfrac{\pi r I B_0}{mg}$.

Units: $[\pi r I B_0] = $ m$\cdot$A$\cdot$T $= $ m$\cdot$A$\cdot$(kg$/$(A$\cdot$s$^2$)) $= $ N. $[mg] = $ N. Ratio dimensionless, consistent with $	an	heta$. Answer (A).

JEE Advanced 2024 Paper 2 · Q04 · Magnetic Field

Solution