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 and radius and it is in a uniform vertical magnetic field , as shown in the figure. [2024]

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 $B_{0}$ , 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 [2024]

Smart Achievers · Accepted Answer

(1)

Let the loop make an angle $θ$ with the vertical.

In equilibrium, $τ_{net} = 0$

$τ_{0} = M B \sin (90 - θ) - m g . r \sin θ = 0$

$I . π r^{2} . B_{0} \cos θ = m g r . \sin θ$

$∴$ $\frac{\sin θ}{\cos θ} = \frac{I π r^{2} B_{0}}{m g r}$

$∴$ $\tan θ = \frac{π r I B_{0}}{m g}$

Solution

1 $\tan θ = \frac{π r I B_{0}}{m g}$

2 $\tan θ = \frac{2 π r I B_{0}}{m g}$

3 $\tan θ = \frac{π r I B_{0}}{2 m g}$

4 $\tan θ = \frac{m g}{π r I B_{0}}$

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Solution

1 tanθ=πrIB0mg

2 tanθ=2πrIB0mg

3 tanθ=πrIB02mg

4 tanθ=mgπrIB0

Ans. (1) Let the loop make an angle θ with the vertical. In equilibrium, τnet=0 τ0=MBsin(90-θ)-mg.rsinθ=0 I.πr2.B0cosθ=mgr.sinθ ∴ sinθcosθ=Iπr2B0mgr ∴ tanθ=πrIB0mg

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1 $\tan θ = \frac{π r I B_{0}}{m g}$

2 $\tan θ = \frac{2 π r I B_{0}}{m g}$

3 $\tan θ = \frac{π r I B_{0}}{2 m g}$

4 $\tan θ = \frac{m g}{π r I B_{0}}$