Question

A circular coil of 160 turns has a radius of 1.92cm. (a)Calculate the current that results in a magnetic dipole moment of magnitude 2.30 Am² . (b)Find the maximum magnitude of the torque that the coil, carrying this current, can experience in a uniform 35.0 mTmagnetic field.

Short answer

1. The value of current is, 12.7 A .
2. The maximum magnitude of the torque is, 0.0805 Nm .

Step-by-step solution

Identification of the given data

1. The number of turns of the coil is, N = 160 .
2. The radius of the coil is, r = 1.90 cm = 0.019 m .
3. The magnetic dipole moment of the coil is, $\mu =2.30A.m^2$.
4. The external magnetic field is, $B=35.0mT=35.0\times {10}^{-3}T$.

Understanding the concept

When a coil carries current, it creates a magnetic dipole moment. We can find the current through the coil by substituting the given values in the formula for magnetic dipole moment. The torque acting on the coil due to external magnetic field is given by the cross product of the magnetic dipole moment of the loop and the external magnetic field.

$$\begin{gathered} \mathit{\mu }\mathbf{=}\mathit{N}\mathit{i}\mathit{A} \\ \mathit{\tau }\mathbf{=}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{\mu }}\mathbf{\times }\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}}\mathbf{=}\mathit{\mu }\mathit{s}\mathit{i}\mathit{n}\mathit{\theta } \end{gathered}$$

(a) Calculate the current that results in a magnetic dipole moment of magnitude 2.30 A.m2

The magnetic moment is calculated as

$\mu =NiA$

For, $A=\pi r^2$,

Then,

$\mu =Ni{\pi }^2$

Rearranging the equation,

$i=\frac{\mu }{N\pi r^2}$

Substitute all the value in the above equation.

$\begin{array}{rcl}i& =& \frac{2.30A{m}^2}{160\times 3.14\times {\left(0.019m\right)}^2}\\ & =& 12.7A\end{array}$

Hence the value of current is, $\begin{array}{rcl}& & 12.7A\end{array}$.

(b) Calculate the maximum magnitude of the torque that the coil

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

The torque is calculated as

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

The magnitude of the torque will be maximum when $\theta =1$, i.e. when $\theta ={90}^o$

$\tau =\mu B$

Substitute all the value in the above equation.

$$\begin{gathered} \tau =2.30A{m}^2\times 35.0\times {10}^{-3}T \\ \tau =0.0805Nm \end{gathered}$$

Hence the maximum magnitude of the torque is, 0.0805Nm .