MCQ
A rectangular coil $PQ$ has $2n$ turns, an area $2a$ and carries a current $2I,$ (refer figure). The plane of the coil is at $60^o$ to a horizontal uniform magnetic field of flux density $B.$ The torque on the coil due to magnetic force is
  • A
    $BnaI ,sin60^o$
  • $8BnaI \,cos60^o$
  • C
    $4naI\, Bsin60^o$
  • D
    none

Answer

Correct option: B.
$8BnaI \,cos60^o$
b
$\vec{\tau}=\vec{M} \times \vec{B}$

Here $\mathrm{M}=2 \mathrm{n}(2 \mathrm{I})(2 \mathrm{a})$

$M=8$ $nI \mathrm{a}$

$\therefore \quad \tau=M B \sin \left(90^{\circ}-60^{\circ}\right)$

$\tau=M B \cos 60^{\circ}$

$\tau=8$ nIa $\cos 60^{\circ}$

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