Question

A current loop, carrying a current of 5.0 A, is in the shape of a right triangle with sides 30, 40, and 50 cm. The loop is in a uniform magnetic field of magnitude 80 mT whose direction is parallel to the current in the 50 cm side of the loop.

1. Find the magnitude of the magnetic dipole moment of the loop.
2. Find the magnitude of the torque on the loop.

Short answer

1. The magnitude of the dipole moment of the loop $\mu =0.30A·m^2$
2. The magnitude of the torque is$\tau =0.024N·m$

Step-by-step solution

Given

1. The dimensions of the triangle = 30 cm, 40 cm and 50 cm
2. The current through the loop = i = 5.0 A
3. The external magnetic field = B = 80 mT = 80 x 10^-3 T
4. The direction of the magnetic field is parallel to the current in the 50 cm side of the triangle.

Determine the concept and the formula for the torque:

The current flowing through a closed wire in the shape of right triangle produces a magnetic moment. It can be calculated using the number of turns of the wire, magnitude of the current, and the area enclosed by the shape. The torque experienced by the shape will be the cross product of the magnetic moment and the external magnetic field.

Consider the formula for the torque as:

$$\begin{gathered} \mu =NiA \\ \tau =\stackrel{\rightharpoonup }{\mu }\times \stackrel{\rightharpoonup }{B}=\mu Bsin\theta \end{gathered}$$

(a) Calculate the magnitude of the magnetic dipole moment of the loop.

The magnetic moment is calculated as,

$\mu =NiA$

Here N = 1 and A = Area of triangle =$\frac{1}{2}\times b\times h$

$\therefore \mu =i\times \left(\frac{1}{2}\times b\times h\right)$

Substitute the values and solve as:

$$\begin{gathered} \mu =5.0\times \frac{1}{2}\times \left(30\times {10}^{-2}\right)\times \left(40\times {10}^{-2}\right) \\ \mu =\frac{0.60}{2} \\ \mu =0.30A·m^2 \end{gathered}$$

(b) Calculate the magnitude of the torque on the loop.

Consider the current carrying loop is placed in an external magnetic field, it experiences a torque.

The torque is calculated as

$$\begin{gathered} \tau =\stackrel{\rightharpoonup }{\mu }\times \stackrel{\rightharpoonup }{B} \\ \therefore \tau =\mu B\sin \theta \end{gathered}$$

The magnetic dipole moment is directed along the vector perpendicular to the plane of the triangle.

Hence, the angle made by magnetic dipole moment with the magnetic field is$\theta ={90}^o$

So the torque experienced is

$\tau =\mu B$

Substitute the values and solve as:

$$\begin{gathered} \tau =0.30\times 80.0\times {10}^{-3} \\ \tau =0.024N·m \end{gathered}$$