Give expression for the force on a current carrying conductor in a magnetic field.

As shown in figure, consider a conductor $PQ$ of length $l$ are of cross-section $A$, carrying current $I$ along + ve $y$-direction. The field $\overrightarrow{\mathrm{B}}$ acts along + ve $z$-direction.

The electrons drift towards left with velocity $\overrightarrow{v_{d}}$.

Each electron experience force along + ve X-axis which is given by,

$\vec{f}=-e\left(\overrightarrow{v_{d}} \times \overrightarrow{\mathrm{B}}\right)$

If $n$ is the number of free electrons per unit volume, then total number of electrons in the conductor is,

$\mathrm{N}=n \times$ Volume $=n \mathrm{Al}$

Total force on the conductor is,

$\overrightarrow{\mathrm{F}}=\mathrm{N} \vec{f}=n \mathrm{Al}\left[-e\left(\overrightarrow{v_{d}} \times \overrightarrow{\mathrm{B}}\right)\right]$

$=n \mathrm{~A} e\left(-\left(l \overrightarrow{v_{d}} \times \overrightarrow{\mathrm{B}}\right)\right)$

But I $\vec{l}$ represents a current element vector in the direction of current so we can take

$\vec{v}_{d}=v_{d} \vec{l}$

$\therefore \quad\overrightarrow{\mathrm{F}}=n \mathrm{Ae}\left(v_{d} \vec{l} \times \overrightarrow{\mathrm{B}}\right)$

$=n \mathrm{Aev}_{d}(\vec{l} \times \overrightarrow{\mathrm{B}})$

but $n$ Aev $_{d}=$ current $I$

$\therefore \vec{F}=\mathrm{I}(\vec{l} \times \overrightarrow{\mathrm{B}})$

and magnitude $\mathrm{F}=\mathrm{I} l \mathrm{~B} \sin \theta$

where $\theta$ is angle between $\overrightarrow{\mathrm{B}}$ and $\mathrm{I}$.

This equation can be applicable for straight conducting rod.

If the wire has an arbitrary shape we can calculate the Lorentz force on it by considering it as a collection of linear strips $d l$ and summing,

$\therefore \overrightarrow{\mathrm{F}}=\sum_{i=1}^{n} \overrightarrow{d l}_{i} \times \overrightarrow{\mathrm{B}}$ where, $i=1,2,3, \ldots, n$