Question

A wire of length 25.0cm carrying a current of 4.51mAis to be formed into a circular coil and placed in a uniform magnetic field$\overrightarrow{B}$of magnitude 5.71mT. If the torque on the coil from the field is maximized. What are (a) the angle between $\overrightarrow{B}$and the coil’s magnetic dipole moment? (b) the number of turns in the coil? (c) What is the magnitude of that maximum torque?

Short answer

1. The angle between $\overrightarrow{B}$and the coil’s magnetic dipole moment is localid="1663952997688" $\theta ={90}^{\circ }$.
2. The number of turns in the coil is N=1.
3. The magnitude of the maximum torque is ${\tau }_{max}=1.28\times {10}^{-7}N.m$.

Step-by-step solution

Given

The length of the wire is L=25.0cm=0.25m.

The current in the coil is i=4.51mA=4.51$\times {10}^{-3}A$.

The magnitude of magnetic field is $B=5.71mT=5.71\times {10}^{-3}T$.

Understanding the concept

By taking cross product of magnetic moment vector and magnetic field vector, we can find the angle between $\overrightarrow{B}$and the coil’s magnetic dipole moment when the torque is maximized. By finding the equation of radius rof coil in the form of length of the wire and substituting it in the formula for themagnetic moment, we can find the number of turns in the coil whenthemagnetic moment ismaximized. We can find the magnitude of the maximum torque by using its formula.

Formula:

The torque is given by$\tau ⃗=\mu ⃗\times B⃗$

The length of wire is$L=N\left(2\pi r\right)$

The magnitude of the magnetic moment is$\mu =NiA$

The magnitude of the maximum torque is given by${\tau }_{max}=\mu B$

(a) Calculate the angle between B→ and the coil’s magnetic dipole moment 

We know that torque is given by

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

The torque is maximum when $\theta ={90}^{\circ }$

Therefore, when the torque is maximized, the angle between $\overrightarrow{B}$and coil’s magnetic dipole moment is $\theta ={90}^{\circ }$

(b) Calculate the number of turns in the coil

The length of wire is$L=N\left(2\pi r\right)$

Where N is the number of turns of the coil,$2\mathrm{\pi r}$is the circumference of the coil.

Therefore, the radius is

$r=\frac{L}{2\pi N}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

The magnitude of the magnetic moment is

$\mu =NiA$

But, area A=${\mathrm{\pi r}}^2$, therefore,

$\mu =Ni\pi r^2$

From equation (1), we get,

$\mu =Ni\pi {\left(\frac{L}{2\mathrm{\pi N}}\right)}^2$

role="math" localid="1662961462302" $\mu =\frac{\left(L^{2}i\right)}{4\mathrm{\pi N}}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

Thus the torque becomes,

$\overrightarrow{\tau }=\mu B=\frac{\left(L^{2}iB\right)}{4\mathrm{\pi N}}$

Thus, to maximize the torque, the number of turns N should be minimum.

So, the number of turns N=1.