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Q.
In using Ampere's law to find the magnetic field of a straight, long solenoid, the loop (Amperian loop) that is taken is
J & K CETJ & K CET 2012Moving Charges and Magnetism
Solution:
When a current carrying coil is placed in a uniform magnetic field $B$, then torque,
$\tau=M B \sin \theta$
$\tau=(N I A) B \sin \theta$
$\tau=10 \times 1 \times 0.2 \times 6 \times 10^{-4} \times \sin 30^{\circ}$
$=10 \times 1 \times 0.2 \times 6 \times 10^{-4} \times \frac{1}{2}$
$=6 \times 10^{-4} N-m$