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
B
IBl
C
32Il2B2
D
34Bl2

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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