Question

Starting from the Lorentz force law, in the form of Eq. 5.16, show that the torque on any steady current distribution (not just a square loop) in a uniform field B is $\mathbf{m}\mathbf{\times }\mathbf{B}\mathbf{.}$

Short answer

Therefore, the torque on any steady current distribution (not just a square loop) in a uniform field B is$m\times B.$

Step-by-step solution

Write the given data from the question.

The Lorentz force law,

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

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

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

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

$dN=r\times dF$

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

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

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

Now,

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

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

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

From equations (2) and (3).

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

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

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

Calculate the total current exerted on the steady current distribution.

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

As,$\oint d\left[r\times (r\times B)\right]=0$ and$\oint B\times \left(r\times d\right)=B\times 2a$

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

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

The magnetic dipole moment is given by,

$m=Ia$

Substitute for into equation (5).

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

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