# What is the magnitude of the torque on the current loop in the figure(Figure 1) ?

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?

(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.

### Raymond Puzio

Raymond Puzio has a PhD in Physics from Yale University. I have been creating PlanetPhysics with Aaron Krowne and Ben Loftin since 2005.