Part A
What is the magnitude of the torque on the current loop in the figure(Figure 1) ?
Part B
What is the loop's equilibrium orientation?
Answer
(A)
The expression for magnetic flux density of the wire is,
B=\frac{\mu I_{n}}{2 \pi r} \sin 90^{\circ}
=\frac{\mu I_{w}}{2 \pi r}
=\frac{\left(4 \pi \times 10^{-7}\right)(2 \mathrm{~A})}{2 \pi(0.02 \mathrm{~m})}
=2 \times 10^{-5} \mathrm{~T}
The expression for torque acting on body is,
\tau=\operatorname{NAB} \sin \theta
The area of the circular loop is,
A=\pi r^{2}
The diameter of the loop is,
d=2.0 \mathrm{~mm}
Now, the radius of the loop is
r=\frac{d}{2}
=\frac{(2.0 \mathrm{~mm})}{2}
=1.0 \times 10^{-1} \mathrm{~m}
Now substitute all the values in the torque equation.
\tau=(1)(0.2 A)(3.14)\left(1.0 \times 10^{-3} \mathrm{~m}\right)^{2}\left(2 \times 10^{-5} \mathrm{~T}\right) \sin \left(90^{\circ}\right)
=1.256 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}
=1.3 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}
(B)
When the loop is rotated to 90^{\circ} then the equilibrium orientation of the loop will be along the horizontal direction.
