Q.Derive the expression for the force on a current-carrying conductor in a magnetic field.

Magnetic Force on Current

Imagine a garden hose spraying pure water — bring a magnet near the stream and nothing happens, because water is electrically neutral. But if that stream carried electric charge (a current), the magnet would push the whole stream sideways. That is the essence of this concept: a current-carrying wire placed in a magnetic field experiences a sideways force, because every moving charge inside the wire feels the Lorentz force, and since the charges cannot leave the wire, they drag the wire along with them.

From a single charge to a wire

A single charge $q$ moving with velocity $\vec v$ in a field $\vec B$ feels $\vec F = q(\vec v \times \vec B)$. A current is just many such charges drifting together, so summing their individual forces over the whole wire gives a net force on it.

The formula

For a straight wire of length $L$ carrying current $I$ in a uniform field $\vec B$:

where $\vec L$ points along the current and $\theta$ is the angle between the wire and $\vec B$.

Direction: the right-hand rule

Point your index finger along the current ($\vec L$), your middle finger along the field ($\vec B$); your thumb then gives the force direction — this is just the cross product $\vec L \times \vec B$ read off by hand. Because it is a cross product, swapping the two vectors reverses the force.

Worked example

A $0.5\ \text{m}$ wire carries $3\ \text{A}$ from east to west, in a uniform field of $0.2\ \text{T}$ pointing north.

• $\theta = 90^\circ$ (the wire and the field are perpendicular), so $F = ILB\sin\theta = (3)(0.5)(0.2)(1) = 0.3\ \text{N}$. …