MCQ
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
  • A
    $Zero$
  • B
    $IBl$
  • C
    $\frac{{\sqrt 3 }}{2}I{l^2}{B^2}$
  • $\frac{{\sqrt 3 }}{4}IB{l^2}$

Answer

Correct option: D.
$\frac{{\sqrt 3 }}{4}IB{l^2}$
d
(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$

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