91Ó°ÊÓ

The Lorentz force law,

$\mathbf{df}\mathbf{=}\mathbf{l}\mathbf{(}\mathbf{dl}\mathbf{×}\mathbf{B}\mathbf{)}$

Here,$I$ is the current,$dl$ is element of the length of the wire and B magnetic field.

The expression for torque due to$dF$on element is given by,

$dN=r×dF$

Here, r is the distance between the axis of rotation and point of application of force.

Substitute $I(dl×B)$for $dF$into above expression.

localid="1657613596027" $$\begin{gathered} dN=r×I(dl×B) \\ dN=lr×(dl×B)............\left(1\right) \end{gathered}$$

Now,

$d\left[r×(r×B)\right]=dr(r×B)+r×(dr×B)$ …… (2)

Substitute $dl$for $dr$into above equation.

$$\begin{gathered} d\left[r×\left(r×B\right)\right]=dl×(r×b)+r×(dl×B) \\ dl×\left(B×r\right)=r×(dl×B)-d\left[r×(r×B)\right].......\left(3\right) \end{gathered}$$

From equations (2) and (3).

$$\begin{gathered} 2r×(dl×B)+B×(r×dl)-\left[r×(r×B)\right]=0 \\ 2r×(dl×B)=d\left[r×(r×B)\right]-B×(r×dl) \\ r×(dl×B)=\frac{1}{2}\left\{d\left[r×(r×B)\right]-B×(r×dl)\right\} \end{gathered}$$

Substitute the equation (3) into equation (1).

$dN=l\frac{1}{2}\left\{d\left[r×(r×B)\right]-B×(r×dl)\right\}$

Calculate the total current exerted on the steady current distribution.

$N=l\frac{1}{2}\left\{âˆ\circledR d\left[r×(r×B)\right]-âˆ\circledR B×(r×dl)\right\}$ …… (4)

As,$âˆ\circledR d\left[r×(r×B)\right]=0$ and$âˆ\circledR B×\left(r×d\right)=B×2a$

Substitute 0 for$âˆ\circledR d\left[r×(r×B)\right]=0$ and$B×2a$ for$âˆ\circledR B×(r×dl)$ into equation (4).

role="math" localid="1657619603868" $$\begin{gathered} N=l\frac{1}{2}\left\{0-B×2a\right\} \\ N=lB×a \\ N=B×al........\left(5\right) \end{gathered}$$

The magnetic dipole moment is given by,

$m=Ia$

Substitute for into equation (5).

$$\begin{gathered} N=-B×m \\ N=m×B \end{gathered}$$

Hence it is shown that the torque on any steady current distribution (not just a square loop) in a uniform field B is $m×B.$