Question

Deduce an expression for the force on a current carrying conductor placed in a magnetic field.

Answer 1

Force on a current-carrying conductor placed in a magnetic field: Let us consider a conductor $PQ$ of length $l$ and area of cross-section $A$. The conductor is placed in a uniform magnetic field of induction $B$ making an angle $\theta$ with the field (figure). A current I flows along $PQ$. Hence, the electrons are drifted along $QP$ with drift velocity $vd$. If $n$ is the number of free electrons per unit volume in the conductor, then the current is
$I=nA{v}_{d}e$
Multiplying both sides by the length of the conductor,
$\therefore Il=nA{v}_{d}el$
Therefore the current element,
$\overrightarrow{Il}=-nA{\overrightarrow{v}}_{d}el....\left(1\right)$
The negative sign in the equation indicates that the direction of current is opposite to the direction of drift velocity of the electrons. Since the electrons move under the influence of magnetic field, the magnetic Lorentz force on a moving electron.
$\overrightarrow{f}=e(\overrightarrow{v_d}\times \overrightarrow{B})....\left(2\right)$
The negative sign indicates that the charge of the electron is negative.
The number of free electrons in the conductor
$N=nAl......\left(3\right)$
The magnetic lorentz force on all the moving free electrons
$\overrightarrow{F}=\overrightarrow{Nf}$,
Substituting equations $\left(2\right)$ and $\left(3\right)$ in the above equation
$\overrightarrow{F}=nAl\{-e(\overrightarrow{v}\times \overrightarrow{b}\left)\right\}$
$\overrightarrow{F}=-nAle\overrightarrow{v}\times \overrightarrow{B}....\left(4\right)$
Substituting equation $\left(1\right)$ in equation $\left(4\right)$
$\overrightarrow{F}=\overrightarrow{Il}\times \overrightarrow{B}$
This total force on all the moving free electrons is the force on the current $t$ carrying conductor placed in the magnetic field.
Magnitude of the force: The magnitude of the force is $F=BIlsin\theta$
(i) If the conductor is placed along the direction of the magnetic field, $\theta =0^{\circ }$, therefore force $F=0$.
(ii) If the conductor is placed perpendicular to the magnetic field, $\theta ={90}^{\circ },F=BIl$. Therefore the conductor experiences maximum force.