Question

Figure shows a rectangular 20-turn coil of wire, of dimensions $\mathbf{10}\mathbf{\text{cm}}$by $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\text{cm}}$. It carries a current of $\mathbf{010}\mathit{A}$and is hinged along one long side. It is mounted in the x-yplane, at angle $\mathbf{\theta }\mathbf{=}{\mathbf{30}}^{\mathbf{0}}$to the direction of a uniform magnetic field of magnitude $\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{\text{T}}$. In unit-vector notation, what is the torque acting on the coil about the hinge line?

Short answer

The torque acting on the coil about the hinge line is $\left(-4.3\times {10}^{-3}\text{N - m}\right)\hat{j}$.

Step-by-step solution

Write the given data

a) Number of turns of coil, $N=20$

b) The length of coil, $l=10\text{cm or}0.10\text{m}$

c) The breadth of coil, $b=5\text{cm or}0.05\text{m}$

d) The current through the coil, $i=0.10\text{A}$

e) The angle between the coil and direction of magnetic field is, $\theta \text{'}=30°$

f) The magnetic field is, $B=0.50\text{T}$

Determine the formula for the torque

The area of the rectangle is as follows:

$A=l\times b$ …… (i)

Here, length of the rectangle, is the breadth of the rectangle.

The torque acting at a point inside a magnetic field is,

$\tau =NiABsin\theta$ …… (ii)

Here, $N$is the number of turns in the coil, $i$is the current of the wire, $A$ is the area of the conductor, $B$is the magnetic field, $\theta$is the angle made by the conductor with the magnetic field.

Determine the torque acting on the coil about the hinge line

The current passing through the segment AD which is along

the hinge line is in the +y direction, and the magnetic field is in the x-z plane. So, the torque acting on the coil will be in the –y direction.

Here,$\theta$is the angle between magnetic dipole moment and the magnetic field.

So, $\theta =90-\theta \text{'}=60°$

Also, the area of rectangular coil can be given using the given data in equation (i) as follows:

$$\begin{gathered} A=0.10\text{m}\times 0.05\text{m} \\ =0.005{\text{m}}^{\text{2}} \end{gathered}$$

Substituting all the known values in the equation (ii), we get the torque value about the hinge line as follows:

$$\begin{gathered} \tau =20\times 0.10\text{A}\times 0.005{\text{m}}^2\times 0.50\text{T}\times \text{sin}\left({60}^0\right) \\ =4.3\times {10}^{-3}\text{N - m} \end{gathered}$$

As, this torque is in the –y direction, we can write it in unit vector notation as,

$\overrightarrow{\tau }=\left(-4.3\times {10}^{-3}\text{N - m}\right)\hat{j}$

Hence, the required value of the torque is $\left(-4.3\times {10}^{-3}\text{N - m}\right)\hat{j}$.