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?

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, $B=0.50\text{T}$

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.

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