Question

Derive an expression for the force on current carrying in a magnetic field.

Answer 1

Consider a small element of conducting wire carrying current $i$ in magnetic field $B$.

The amount of charge in a small element is given as,

$dq=idt$

The Lorenz force is given as,

$d\overrightarrow{F}=dq\left(\overrightarrow{v}\times \overrightarrow{B}\right)$

$d\overrightarrow{F}=idt\left(\overrightarrow{v}\times \overrightarrow{B}\right)$

$d\overrightarrow{F}=i\left(\left(\overrightarrow{v}dt\right)\times \overrightarrow{B}\right)$

$d\overrightarrow{F}=i\left(\left(\overrightarrow{d}l\right)\times \overrightarrow{B}\right)$

If the magnetic field is constant then,

$\int d\overrightarrow{F}=i\left(\left(\int \overrightarrow{d}l\right)\times \overrightarrow{B}\right)$

$\overrightarrow{F}=i\left(\overrightarrow{L}\times \overrightarrow{B}\right)$

If the conductor is a straight wire and the length is $l$, then the magnitude of force is given as,

$F=ilB\sin \theta$

Where, the angle between the current direction in the wire and the direction of the magnetic field is $\theta$.

The direction of the magnetic field is determined by Fleming's left hand rule.