Calculate force on a 2m wire carrying 5A in 0.3T field.

Magnetic Force on a Current-Carrying Wire — F = BIL

Question

A straight wire of length 2 m carries a current of 5 A. It is placed in a uniform magnetic field of strength 0.3 T, with the wire perpendicular to the field. Find the magnetic force acting on the wire.

Solution — Step by Step

Length of wire: $L = 2\ \text{m}$ , Current: $I = 5\ \text{A}$ , Magnetic field: $B = 0.3\ \text{T}$ , Angle between wire and field: $\theta = 90°$ .

The problem says “perpendicular to the field” — that’s your cue that $\sin\theta = 1$ , which gives maximum force.

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

F = BIL\sin\theta

This formula comes from the Lorentz force law acting on every moving charge in the wire, summed over the entire length. We’ll see why in the next section.

F = 0.3 \times 5 \times 2 \times \sin 90°

F = 0.3 \times 5 \times 2 \times 1

F = 3\ \text{N}

Point your index finger along $\vec{B}$ , middle finger along the direction of current $I$ — your thumb points in the direction of the force. Always state direction in CBSE and JEE descriptive answers; numerical value alone loses marks.

The magnetic force on the wire is $F = 3\ \text{N}$ .

Why This Works

Every charge carrier inside the wire experiences a Lorentz force $\vec{F} = q\vec{v} \times \vec{B}$ . Since current $I = nqv_dA$ (where $n$ is charge carrier density, $v_d$ is drift velocity, $A$ is cross-section area), when you sum this force over all carriers across length $L$ , it neatly collapses to $F = BIL\sin\theta$ .

The $\sin\theta$ factor tells us force is maximum when the wire is perpendicular to $\vec{B}$ and zero when the wire is parallel to $\vec{B}$ . Think about it physically: if the wire runs along the field, the drift velocity of charges is also along $\vec{B}$ , and $\vec{v} \times \vec{B} = 0$ .

This is a high-weightage concept in CBSE Class 12 Chapter 4 and regularly appears in JEE Main as a one-mark or two-mark numerical. The formula is simple — the marks come from handling the angle correctly.

Alternative Method — Using Vector Form

If the wire direction and field direction are given as vectors, use:

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

Say current flows along $\hat{i}$ (x-axis) and $\vec{B} = 0.3\hat{j}$ :

\vec{F} = 5 \times (2\hat{i}) \times (0.3\hat{j}) = 5 \times 0.6\ (\hat{i} \times \hat{j}) = 3\hat{k}\ \text{N}

Force is along $+z$ direction. This vector method is essential for JEE Main problems where direction matters and they give you unit vector notation.

In JEE Main MCQs, when two options have the same magnitude but different directions, the vector cross-product method is the fastest way to nail the direction without ambiguity.

Common Mistake

Students forget to check the angle and directly plug in $F = BIL$ without the $\sin\theta$ factor. This works here because $\theta = 90°$ and $\sin 90° = 1$ . But if the problem says the wire makes a 30° or 60° angle with the field, you must include $\sin\theta$ . A typical trap question: “wire makes 30° with $\vec{B}$ ” — many students write $F = BIL$ and lose full marks. Always write the complete formula first, then substitute.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Vector Form

Common Mistake

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