Derive an expression for magnetic force acting on a current carrying conductor placed in uniform magnetic field.
A current carrying conductor contains large number of free electrons.These electrons move with drift velocity $\vec{v}_d$ in a direction opposite to the direction of conventional current flowing in the conductor. An electron moving in a uniform magnetic field $\vec{B}$ experiences a force which is transmitted to the conductor.
Consider a conductor of length $l$ carrying a current $I$ placed in a uniform magnetic field $\vec{B}$.
Let $n$ = Number of free electrons per unit volume of the conductor.
Let $A$ = Area of cross-section of the conductor.
Magnetic force acting on an electron is given by:
$\vec{f}_m = -e (\vec{v}_d \times \vec{B}) \quad \dots (i)$
Now consider a small element of length $d\vec{l}$ of the given conductor.
Number of electrons in the small element = n × volume of the element = $n (A dl)$.
Magnetic force experienced by the element of the conductor is given by:
$d\vec{F}_m$ = number of electrons in the element × force on each electron
$d\vec{F}_m = (n A dl)\vec{f}_m$
$d\vec{F}_m = (n A dl) [-e (\vec{v}_d \times \vec{B})]$
$d\vec{F}_m = -n A e dl (\vec{v}_d \times \vec{B}) \quad \dots (ii)$
But drift velocity,
$\vec{v}_d = -\frac{d\vec{l}}{dt}$
( $\because \vec{dl}$ is in a direction opposite to the direction of $\vec{v}$)
$(nA dl) e = dq$
where, $dq$ is the charge flowing in the small element.
Now
$d\vec{F}_m = dq (\frac{d\vec{l}}{dt} \times \vec{B})$
$d\vec{F}_m= \frac{dq}{dt}(d\vec{l} \times \vec{B}) \quad \dots (iii)$
Now , $\frac{dq}{dt} = I$
$\therefore$ , $d\vec{F}_m=I (d\vec{l} \times \vec{B})$
Since conductor is made of large number of such elements, therefore, total force experienced by the conductor is given by,
$\vec{F}_m= \int d\vec{F}_m = \int I (d\vec{l} \times \vec{B})$
$(\because \int \vec{dl} = \vec{l})$
$\vec{F}_m = I ( \int d\vec{l} \times \vec{B})$
$\vec{F}_m = I (\vec{l} \times \vec{B}) \quad \dots (iv)$
Magnitude of $\vec{F}_m$ is given by
$F_m = BIl \sin \theta$
where, $\theta$ is the angle between $\vec{l}$ and $\vec{B}$.
Special Cases:
(i) If $\theta = 0^\circ$ or $180^\circ$, $F_m = 0$. It means current carrying conductor experiences no force when placed parallel or anti-parallel to the direction of the magnetic field.
(ii) If $\theta = 90^\circ$, $F_m = BIl$. It means a current carrying conductor placed at right angle to the uniform magnetic field experiences maximum force.
Direction of $\vec{F}_m$ is perpendicular to the plane containing $\vec{l}$ and $\vec{B}$.
The direction of $\vec{F}_m$ is determined by Fleming's left hand rule.
