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Physics Question on Uniform Circular Motion

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 B0B_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
Alternative_text

A

tanθ=πrlB0(𝑚𝑔)tan\theta = \frac{\pi rlB_0}{(𝑚𝑔)}

B

tanθ=2πrlB0(𝑚𝑔)tan\theta = \frac{2\pi rlB_0}{(𝑚𝑔)}

C

tanθ=πrlB0(2𝑚𝑔)tan\theta = \frac{\pi rlB_0}{(2𝑚𝑔)}

D

tanθ=mg(πrlB0)tan\theta = \frac{mg}{(\pi rlB_0)}

Answer

tanθ=πrlB0(𝑚𝑔)tan\theta = \frac{\pi rlB_0}{(𝑚𝑔)}

Explanation

Solution

The correct option is (A):tanθ=πrlB0(𝑚𝑔)tan\theta = \frac{\pi rlB_0}{(𝑚𝑔)}