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Question

A triangular loop of side $$l$$ carries a current $$I$$. It is placed in a magnetic field $$B$$ such that the plane of the loop in the direction of $$B$$. The torque on the loop is:


A
Zero
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B
IBl
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C
32Il2B2
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D
34Bl2
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Solution

The correct option is D $$\dfrac{\sqrt 3}{4} B l^2$$
Since $$q = {90^0}$$
So$$,$$ $$\tau  = NIAB = 1 \times I \times \left( {\frac{{\sqrt 3 }}{4}{l^2}} \right)B = \frac{{\sqrt 3 }}{4}{l^2}B$$
Hence,
option $$(D)$$ is correct answer.

Physics

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