F/L = μ₀I₁I₂/(2πd). Attract if same direction, repel if opposite.

Force Between Two Parallel Current-Carrying Wires

Question

Two long parallel wires are separated by a distance of 0.1 m. Wire 1 carries a current of 10 A and Wire 2 carries a current of 5 A, both in the same direction. Find the force per unit length between them and state whether they attract or repel.

Solution — Step by Step

The force per unit length between two parallel current-carrying wires is:

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

Here $\mu_0 = 4\pi \times 10^{-7}$ T·m/A, $d$ is the separation between the wires. This formula comes directly from the Biot-Savart field of one wire acting on the other — we’ll see why in the “Why This Works” section.

We have $I_1 = 10$ A, $I_2 = 5$ A, $d = 0.1$ m.

\frac{F}{L} = \frac{(4\pi \times 10^{-7})(10)(5)}{2\pi \times 0.1}

Notice the $\pi$ in numerator and denominator cancel immediately — always look for this before punching numbers.

After cancelling $\pi$ :

\frac{F}{L} = \frac{4 \times 10^{-7} \times 10 \times 5}{2 \times 0.1} = \frac{4 \times 10^{-7} \times 50}{0.2}

\frac{F}{L} = \frac{200 \times 10^{-7}}{0.2} = 1000 \times 10^{-7} = 10^{-4} \text{ N/m}

Since both currents flow in the same direction, the wires attract each other.

The right-hand rule tells us why: Wire 1’s magnetic field points into the page at Wire 2’s location. Force on Wire 2 is $I_2 \vec{L} \times \vec{B}$ , which points toward Wire 1.

Final Answer: $F/L = 10^{-4}$ N/m, attractive

Why This Works

Wire 1 creates a magnetic field at the location of Wire 2. By Biot-Savart (or Ampere’s law — much faster), that field at distance $d$ is $B_1 = \mu_0 I_1 / (2\pi d)$ , circling the wire.

Wire 2, carrying current $I_2$ , sits inside this field. The force on a current-carrying conductor in a magnetic field is $F = BIL$ , giving $F/L = B_1 I_2 = \mu_0 I_1 I_2 / (2\pi d)$ .

The direction rule is the part students actually need to memorise for CBSE and JEE: same direction → attract, opposite direction → repel. This is literally the opposite of parallel charges (same sign repel), so your brain will try to flip it. Don’t let it.

Alternative Method — Using Field Superposition Thinking

Instead of memorising the formula, rebuild it on the spot during the exam:

Step 1: $B$ due to a long straight wire at distance $d$ :

B = \frac{\mu_0 I}{2\pi d}

Step 2: Force on Wire 2 of length $L$ in this field:

F = I_2 L B = I_2 L \cdot \frac{\mu_0 I_1}{2\pi d}

Step 3: Divide both sides by $L$ :

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

This two-step derivation takes under 30 seconds and is worth doing in JEE Main if you blank on the formula. The examiners award method marks.

In JEE Main 2023 (January Session), a variant of this problem gave numerical values and asked for the ratio of forces when one current was doubled. The formula shows $F/L \propto I_1 I_2$ , so doubling $I_1$ doubles the force. No recalculation needed — just read the proportionality.

Common Mistake

Forgetting to cancel $\pi$ before computing.

Students substitute $\mu_0 = 4\pi \times 10^{-7}$ and then divide by $2\pi d$ , but multiply everything out numerically instead of cancelling. This leads to a mess of $\pi$ values and arithmetic errors. Always cancel $\pi$ from numerator and denominator first — you’ll almost always get a clean $\frac{2 \times 10^{-7} I_1 I_2}{d}$ form, which is much easier to compute.

The cleaner form: $\displaystyle\frac{F}{L} = \frac{2 \times 10^{-7} \times I_1 I_2}{d}$ . Keep this in your formula sheet.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Field Superposition Thinking

Common Mistake

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