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

The force on a current-carrying conductor in a magnetic field is obtained by summing the magnetic (Lorentz) forces on all the free charge carriers drifting through it, giving $\vec F = I(\vec l \times \vec B)$, i.e. $F = BIl\sin\theta$.

Setup

Consider a straight conductor of length $l$ and uniform cross-sectional area $A$, carrying a steady current $I$, placed in a uniform magnetic field $\vec B$ making angle $\theta$ with the conductor. Let $n$ be the number density of free electrons in the conductor and $v_d$ their drift speed.

Step 1: Force on a single charge carrier

Each free electron (charge magnitude $e$) drifting with velocity $\vec v_d$ in the field $\vec B$ experiences a magnetic (Lorentz) force

$\vec f = -e(\vec v_d \times \vec B)$

The sign here only fixes the direction; the conventional current direction is opposite to the electron drift direction, which is accounted for in the next steps.

Step 2: Number of charge carriers in the conductor

Volume of the conductor $= A l$, so the number of free electrons in it is

$N = nAl$

Step 3: Total force

Since all carriers drift with the same $\vec v_d$, the total magnetic force on the conductor is the sum of the forces on all $N$ carriers:

$\vec F = nAl\, e\,(\vec v_d \times \vec B)$

taking the current direction (conventional current, opposite to electron drift) along the direction of the conductor.

Step 4: Introduce the current

The conventional current is related to the drift speed by

$I = nAev_d$

so $nAe\,v_d = I$. Writing the length of the conductor as a vector $\vec l$ along the direction of current flow (magnitude $l$),

$\vec F = I(\vec l \times \vec B)$

Step 5: Magnitude and direction

$F = BIl\sin\theta$ …