Question

A circular wire loop of radius15.0cmcarries a current of 2.60 A. It is placed so that the normal to its plane makes an angle of 41.0° with a uniform magnetic field of magnitude 12.0 T. (a) Calculate the magnitude of the magnetic dipole moment of the loop. (b) What is the magnitude of the torque acting on the loop?

Short answer

1. $\mu =0.184A·m^2$
   
2. $\tau =1.45N·m$
   

Step-by-step solution

Given

The radius of the wire loop = $r=15.0cm=0.15m$

The current through the loop = $i=2.60A$

The angle between the normal to the plane of the loop and the external magnetic field = $\theta =41.0^o$

The external magnetic field =$B=12.0T$

Understanding the concept

The magnetic dipole moment of a current carrying loop depends on the number of loops, the current through the loop, and the area enclosed by each turn of the loop. Substituting these values in the formula, we can find the magnitude of the magnetic dipole moment. The torque acting on the loop due to external magnetic field is given by the cross product of the magnetic moment of the loop and the external magnetic field.

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

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

The magnetic moment is given by

$\mu =NiA$

Here,$N=1andA=\pi r^2$

$$\begin{gathered} \mu =i\pi r^2 \\ \mu =2.60\times 3.14\times \left(0.{15}^2\right) \\ \mu =2.60\times 3.14\times \left(0.0225\right) \\ \mu =0.184A.m^2 \end{gathered}$$

Hence,$\mu =0.184A.m^2$

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

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

The magnitude of the torque is calculated as

$$\begin{gathered} \stackrel{\rightharpoonup }{\tau }=\stackrel{\rightharpoonup }{\mu }\times \stackrel{\rightharpoonup }{B} \\ \tau =\mu B\sin \theta \\ \tau =0.184\times 12.0\times \sin \left(41.0\right) \\ \tau =0.184\times 12.0\times 0.6561 \\ \tau =1.45N·m \end{gathered}$$

Hence,$\tau =1.45N·m$