JEE Advanced 2024 — Paper 2 — PHY-I solution

JEE Advanced2024Paper 2PHY-I

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 $B_{0}$ , 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 $P Q$ by an angle $θ$ given by

A
$tan θ = π r I B_{0} / (m g)$
B
$tan θ = 2 π r I B_{0} / (m g)$
C
$tan θ = π r I B_{0} / (2 m g)$
D
$tan θ = m g / (π r I B_{0})$

Solution

The current loop has magnetic moment $M = I A = I π r^{2}$ , directed perpendicular to the plane of the loop.

When the loop tilts by angle $θ$ from the vertical plane, its magnetic moment makes angle $(9 0^{\circ} - θ)$ with the vertical field $B_{0}$ .

Magnetic torque about the axis $P Q$ : $τ_{B} = M B_{0} cos θ = I π r^{2} B_{0} cos θ .$

Gravitational restoring torque about $P Q$ : the centre of mass of the loop is at distance $r$ below the pivot, and when tilted by $θ$ , $τ_{g} = m g r sin θ .$

For equilibrium, these balance:

$I π r^{2} B_{0} cos θ = m g r sin θ$

$\Rightarrow tan θ = \frac{π r I B _{0}}{m g} .$

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