Force on current-carrying conductor in magnetic field — F = BIL sin θ

Question

A straight conductor of length 50 cm carries a current of 2 A. It is placed in a uniform magnetic field of 0.5 T at an angle of 30° to the field. Find the magnitude and direction of the force on the conductor.

(NCERT Class 12, Chapter 4)

Solution — Step by Step

The force on a current-carrying conductor in a magnetic field is:

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

The magnitude is: $F = BIL\sin\theta$

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

$B = 0.5$ T, $I = 2$ A, $L = 0.50$ m, $\theta = 30°$

F = 0.5 \times 2 \times 0.50 \times \sin 30°

F = 0.5 \times 2 \times 0.50 \times 0.5

\boxed{F = 0.25 \text{ N}}

Point your index finger along the magnetic field $\vec{B}$ , middle finger along the current $I$ — your thumb points in the direction of the force $\vec{F}$ .

The force is perpendicular to the plane containing $\vec{L}$ and $\vec{B}$ . Its direction is given by the cross product $\vec{L} \times \vec{B}$ .

Why This Works

Each charge carrier (electron) in the conductor experiences a Lorentz force $\vec{f} = q\vec{v}_d \times \vec{B}$ . When we sum over all carriers in the conductor, the drift velocity $v_d$ , charge $q$ , number density $n$ , and cross-section $A$ combine to give: $F = nqv_dAL \cdot B\sin\theta = BIL\sin\theta$ (since $I = nqv_dA$ ).

The force is maximum when the conductor is perpendicular to the field ( $\theta = 90°$ ) and zero when parallel ( $\theta = 0°$ ). This makes physical sense: when the charges move parallel to $\vec{B}$ , the cross product is zero — no magnetic force.

Alternative Method

In vector form: $\vec{F} = I\vec{L} \times \vec{B}$ . If $\vec{L} = L(\cos 30°\,\hat{i} + \sin 30°\,\hat{j})$ and $\vec{B} = B\hat{i}$ , compute the cross product directly:

\vec{F} = IL\sin 30° \cdot B\,(\hat{j} \times \hat{i}) = -ILB\sin 30°\,\hat{k}

Magnitude: 0.25 N, direction: $-\hat{k}$ (into the page, if the current-field plane is the page).

For NEET MCQs, remember the special cases: $F = BIL$ when $\theta = 90°$ and $F = 0$ when $\theta = 0°$ . Most problems use one of these two cases. Also, the force is always perpendicular to both $I$ and $B$ — it can never do work on the conductor (though it can do work on the moving conductor in motional EMF problems).

Common Mistake

Students often write $F = BIL$ without the $\sin\theta$ factor. This only works when the conductor is perpendicular to the field. Always check the angle. Also, some students confuse Fleming’s left-hand rule (for force on a conductor) with the right-hand rule (for field due to a current). Left hand = motor effect (force), right hand = generator effect (or field direction). Mixing them up flips the force direction.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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