**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.