JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Magnetic Effects of Current

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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 PQ by an angle

θ

given by

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

Answer: (A)

Solution

The magnetic moment of the circular loop is

M = I \cdot π r^{2}

. In the vertical magnetic field

B_{0}

, when the loop makes angle

θ

with the vertical, the magnetic torque is

τ_{mag} = M B_{0} cos θ = I π r^{2} B_{0} cos θ

.

The gravitational torque about the axis PQ is

τ_{grav} = m g r sin θ

.

For equilibrium:

I π r^{2} B_{0} cos θ = m g r sin θ

Therefore:

tan θ = \frac{π r I B _{0}}{m g}

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Single correct option

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JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Magnetic Effects of Current

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