\(A=0.01 \mathrm{~m}^2, \mathrm{I}=10 \mathrm{~A}, \mathrm{~B}=0.1 \mathrm{~T},\)
\(\theta=0^{\circ}\)
To find:
Torque acting on the loop, \(\tau\)
Solution:
Magnetic moment, \(M=(I)(A)\)
The torque acting on the loop is given as, \(\vec{\tau}=\vec{M} \times \vec{B}\)
\(\tau=M B \sin \theta \)
\(\tau=1 A B \sin 0\)
\(\tau=0\;\mathrm{Nm}\)
$\left[{m}_{{p}}=1.67 \times 10^{-27} {kg}, {e}=1.6 \times 10^{-19} {C},\right.$ પ્રકાશની ઝડપ $\left.=3 \times 10^{8} {m} / {s}\right]$