91Ó°ÊÓ

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.

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

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.

$M_1=I{\mathrm{Ï€²¹}}^2\hat{\mathrm{z}}$

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

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

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

Substitute $l{\mathrm{Ï€²¹}}^2\hat{\mathrm{z}}$for$M_1$ into above equation.

$$\begin{gathered} B_1=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{1}{r^3}\left[3\left(l{\mathrm{Ï€²¹}}^2\hat{\mathrm{z}}.\mathrm{y}\right)y-m_1\hat{z}\right] \\ {B}_1=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{l{\mathrm{Ï€²¹}}^2}{r^3}\hat{z} \end{gathered}$$

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

$N=M_2×B_1$

Substitute $-\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{l{\mathrm{Ï€²¹}}^2}{r^3}\hat{z}$for $B_1$and $l{b}^{2}y$for $M_2$into above equation.

role="math" localid="1657687839725" $$\begin{gathered} N=l{b}^{2}y×\left(-\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{l{\mathrm{Ï€²¹}}^2}{r^3}\hat{z}\right) \\ N=-\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{l{\mathrm{Ï€²¹}}^2×l{b}^2}{r^3}\hat{y}×\hat{z} \end{gathered}$$

Substitute for into above equation.

role="math" localid="1657687877416" $N=-\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{l{\mathrm{Ï€²¹}}^2×l{b}^2}{r^3}\hat{x}$

Hence, the torque on the square loop due to circular loop is $\frac{{\mathrm{μ}}_0}{4}\frac{l^2{a}^2{b}^2}{r^3}$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.