Calculate the torque exerted on the square loop shown in Fig. 6.6, due to the circular loop (assume is much larger than or ). If the square loop is free to rotate, what will its equilibrium orientation be?

The distance between the square loop and circular loop is r .

The distance r is much larger than or b.

Here, a is the radius of the circular loop and b is the side of the square loop.

The equation to calculate the torque is given as follows.

$\mathbf{N}\mathbf{=}\mathbf{m}\mathbf{\times }\mathbf{B}$

The equation calculate the magnetic dipole moment to is given as follows.

$\mathbf{m}\mathbf{=}\mathbf{IA}$

Here lis the current and Ais the area.

The dot product of two different vector is equal to zero.

$\hat{\mathbf{y}}\mathbf{.}\hat{\mathbf{z}}\mathbf{=}\mathbf{0}$

Calculate the magnetic moment of the circular loop.

$$\begin{gathered} M_1=m_1\hat{z} \\ {M}_1=I{A}_1\hat{z} \end{gathered}$$

Here $A_1$is the area of the circular loop and I is the current.

$M_1=I{\mathrm{\pi a}}^2\hat{\mathrm{z}}$

Calculate the magnetic moment of the square loop.

$$\begin{gathered} M_2=m_{2}y \\ {M}_2=I{A}_{2}y \end{gathered}$$

Here $A_2$is the area of the circular loop,

$M_2=I{b}^{2}y$

Calculate the magnetic strength due to circular loop.

role="math" localid="1657686888944" $B_1=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{1}{r^3}\left[3\left(M_1.\hat{r}\right)\hat{r}-M_1\right]$

Substitute y for $\hat{r}$into above equation.

$B_1=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{1}{r^3}\left[3\left(M_1.\hat{y}\right)\hat{y}-M_1\right]$

Substitute $l{\mathrm{\pi a}}^2\hat{\mathrm{z}}$for$M_1$ into above equation.

$$\begin{gathered} B_1=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{1}{r^3}\left[3\left(l{\mathrm{\pi a}}^2\hat{\mathrm{z}}.\mathrm{y}\right)y-m_1\hat{z}\right] \\ {B}_1=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{l{\mathrm{\pi a}}^2}{r^3}\hat{z} \end{gathered}$$

Calculate the torque exerted on the square loop due to circular loop.

$N=M_2\times B_1$

Substitute $-\frac{{\mu }_0}{4\mathrm{\pi }}\frac{l{\mathrm{\pi a}}^2}{r^3}\hat{z}$for $B_1$and $l{b}^{2}y$for $M_2$into above equation.

role="math" localid="1657687839725" $$\begin{gathered} N=l{b}^{2}y\times \left(-\frac{{\mu }_0}{4\mathrm{\pi }}\frac{l{\mathrm{\pi a}}^2}{r^3}\hat{z}\right) \\ N=-\frac{{\mu }_0}{4\mathrm{\pi }}\frac{l{\mathrm{\pi a}}^2\times l{b}^2}{r^3}\hat{y}\times \hat{z} \end{gathered}$$

Substitute for into above equation.

role="math" localid="1657687877416" $N=-\frac{{\mu }_0}{4\mathrm{\pi }}\frac{l{\mathrm{\pi a}}^2\times l{b}^2}{r^3}\hat{x}$

Hence, the torque on the square loop due to circular loop is $\frac{{\mu }_0}{4}\frac{l^2{a}^2{b}^2}{r^3}$and the square loop to be in the equilibrium the net torque on the loop should be zero therefore, the orientation of the square loop should be in downward direction.