Force on current carrying conductor in magnetic field — derive

Question

Derive an expression for the force on a current-carrying conductor placed in a uniform magnetic field. Under what conditions is this force maximum and zero?

Solution — Step by Step

We know the force on a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is:

\vec{F} = q(\vec{v} \times \vec{B})

A current-carrying conductor is essentially a collection of such moving charges (electrons), so we’ll build up from this single-charge picture.

Take a small element of the conductor of length $d\vec{l}$ carrying current $I$ . Let $n$ be the number density of free electrons (charge carriers), $A$ the cross-sectional area, and $v_d$ the drift velocity of electrons.

The number of charge carriers in the element $dl$ is:

dN = n \cdot A \cdot dl

Each electron has charge $q = e$ (magnitude) and velocity $\vec{v}_d$ . Force on the element:

d\vec{F} = (dN) \cdot e(\vec{v}_d \times \vec{B}) = nA \, dl \cdot e(\vec{v}_d \times \vec{B})

Now recall that current $I = nAev_d$ , so $nAe \cdot v_d = I$ . We can write $e \cdot v_d \cdot dl = I \cdot dl$ directionally:

d\vec{F} = I \, d\vec{l} \times \vec{B}

For a straight conductor of length $L$ in a uniform field:

\vec{F} = I \vec{L} \times \vec{B}

In magnitude:

\boxed{F = BIL\sin\theta}

where $\theta$ is the angle between the direction of current ( $\vec{L}$ ) and the magnetic field ( $\vec{B}$ ).

Maximum force: $\theta = 90°$ , i.e., conductor is perpendicular to the field. $F_{max} = BIL$
Zero force: $\theta = 0°$ or $180°$ , i.e., conductor is parallel to the field. $F = 0$

The direction of force is given by Fleming’s Left Hand Rule: point the forefinger in the direction of $\vec{B}$ , the middle finger in the direction of current $I$ , and the thumb points in the direction of force.

Why This Works

The derivation shows that a current is just many moving charges bunched together. The magnetic force on each carrier adds up, and since current $I = nAev_d$ , all the microscopic details collapse into a clean macroscopic formula $F = BIL\sin\theta$ .

The cross product $d\vec{l} \times \vec{B}$ captures both the magnitude (via $\sin\theta$ ) and the direction (perpendicular to both). When the conductor is parallel to $\vec{B}$ , $\sin 0° = 0$ — the charges move along the field, so the field exerts no sideways push.

JEE Main frequently tests: a current loop in a magnetic field, or a conductor at an angle. Always resolve the current direction relative to $\vec{B}$ first. The formula $F = BIL\sin\theta$ applies only to a straight conductor in a uniform field.

Alternative Method

Using vector notation directly: $\vec{F} = I(\vec{L} \times \vec{B})$ . The magnitude is $|I||L||B|\sin\theta$ and direction follows the right-hand rule for the cross product (or Fleming’s left-hand rule for conventional current).

For a curved conductor in a non-uniform field, we must integrate: $\vec{F} = I \int d\vec{l} \times \vec{B}$ .

Common Mistake

Students often confuse Fleming’s Left-Hand Rule (for force on conductor carrying current — motor principle) with Fleming’s Right-Hand Rule (for EMF induced in a conductor moving in a field — generator principle). Left Hand = current in, force out. Right Hand = force in, EMF/current out.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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