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 is in the direction of $B$. The torque on the loop is
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(d) Since $\theta$ = $ 90°$
Hence $\tau = NIAB = 1 \times I \times \left( {\frac{{\sqrt 3 }}{4}{l^2}} \right)\,B$
$ = \frac{{\sqrt 3 }}{4}{l^2}B$
Hence $\tau = NIAB = 1 \times I \times \left( {\frac{{\sqrt 3 }}{4}{l^2}} \right)\,B$
$ = \frac{{\sqrt 3 }}{4}{l^2}B$
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