[NOTES/EM-07006]-Lorentz Force on a Current Distribution

1. Magnetic force on a current

We will compute the force due to magnetic field on a conductor carrying current density $\vec{j}(\vec{r})$. Let $N$ charges per unit volume carrying charge $q$ be present in the conductor. Then the force on a single charge will be $q\vec{v}\times\vec{B}$ the force on a small volume element $dV$ will be \begin{align} (N~dV)q\vec{v}\times \vec{B} = p\vec{v}\times\vec{B}\,dV\,.\label{eq001} \end{align} Remembering $\vec{j}=\rho\vec{v}$, for a conductor carrying a current we get \begin{align}\label{eq002} \boxed{ \text{force per unit volume} = (\vec{j}\times\vec{B})}\,\,. \end{align}

2. Force on a current in a thin wire Let a current $I$ be flowing in a thin wire of cross section $A$.

Considering the flow of charges in a small line element of length $dl$ and volume $dV=Adl$, the current density $\vec{j}$ is seen to be related to the current $I$ by $\vec{j}dV=\vec{j}Adl=I\overrightarrow{dr}$. This result can be verified by checking that both the quantities have the same magnitude and direction. For an infinitesimal length $dl$ of a wire carrying current $I$, the magnetic force is given by $=\vec{j}\times\vec{B}\,dV=I(A\,\overrightarrow{dr}\times\vec{B}).$ Therefore, for a wire the force can be computed by integrating \begin{align}\label{eq003} \boxed{\text{force on line element} ~dl=I\,\overrightarrow{dr}\times \vec{B}}\,\,. \end{align} It must be remembered that, in our notation, $|\overrightarrow{dr}|=dl$ and that the direction of $\overrightarrow{dr}$ is along the tangent pointing towards the flow of the current.

3.Magnetic force on a current carrying loop

The magnetic force on a line element a wire carrying current $I$ is given by $I\,\overrightarrow{dr}\times\vec{B}$. Therefore, the force experienced by a loop in magnetic field will be given by \begin{align} \vec{F} = \oint I\,\overrightarrow{dr} \times\vec{B}\label{eq004} \end{align} We will show that the force vanishes when the magnetic field is uniform. For this purpose we take the component of the force (\ref{eq004}) along a unit vector $\hat{n}$ and we will prove that $\hat{n}\cdot\vec{F}=0$ for all $\hat{n}$ implying $\vec{F}=0$. Therefore, \begin{align} \hat{n}\cdot\vec{F} = & I\oint \hat{n}\cdot\overrightarrow{dr} \times \vec{B}\label{eq005}\\ =&I\oint (\hat{n}\times \overrightarrow{dr})\cdot\vec{B}\label{eq006}\\ =& I\hat{B}\cdot\oint(\hat{n}\times\overrightarrow{dr})\label{eq007} \end{align} In the last step the magnetic field has been pulled out of the integral sign because it is uniform, i.e., is same at all points of the loop. To show $\oint(\hat{n}\times\overrightarrow{dr})=0$, we its consider $x$-component \begin{align} \oint(\hat{n}\times\overrightarrow{dr}) = & \oint(n_y dz-n_zdy)\label{eq008}\\ =& n_y \oint dz - n_z \oint dy\label{eq009} \end{align} Both the terms vanish because for a closed loop $\oint dx=\oint dy =\oint dz =0$. 

• Remember that :The magnetic force experienced by a current carrying loop in a uniform magnetic field is zero.

References 

QFY 1. For a conducting wire end points $A,B$, and of any shape, the magnetic force in a uniform magnetic field is equal to that on a straight wire having the same end points.

QFY 2. Since magnetic field is uniform and if $\hat{n}$ is a fixed vector, $\nabla\times(\hat{n}\times B)=0$. Use this and Stokes theorem to give an alternate proof of the result that the force on a current loop in uniform field is zero.

QFY 3  What can you say about the force on a current carrying conductor in electric field?

• Sec 13-3  The magnetic force on a current R. P. Feynman, Robert B. Leighton and Mathew Sands Lectures on Physics, vol-II, B.I. Publications (1964)
• Sec 33-3 Magnetic force on a current David Halliday and R. Resnick,  Physics Vol-II, Second Edition, (Revised Printing 1966), New Age International Publishers, New Delhi,(1966).