Attraction, Repulsion and Equilibrium of Parallel Wires

Physics · Moving Charges And Magnetism · NEET

Two long parallel wires carrying currents in the SAME direction attract each other; currents in OPPOSITE directions repel. The force per unit length is F/L = mu_0 * I1 * I2 / (2*pi*d). Memory hook: "Same way = come together, opposite way = push apart."
Same direction: ATTRACTOpposite direction: REPELI1 upI2 upI1 upI2 downF/L = mu_0 I1 I2 / (2 pi d)
Red arrows show the force direction. Left: same-direction currents pull the wires together (attraction). Right: opposite-direction currents push them apart (repulsion). The magnitude follows F/L = mu_0 * I1 * I2 / (2 * pi * d).

Your doubts, answered

Same direction currents: attract or repel? Why do I always mix it up?

Same direction attracts, opposite direction repels. This is the reverse of what most students expect (magnets: like poles repel). The reason: wire A's field at wire B, combined with F = I L x B, points toward A when currents are parallel. Memory hook: parallel currents are 'friends who come together', anti-parallel currents 'push apart'.

What is the force per unit length between two parallel wires?

F/L = mu_0 * I1 * I2 / (2*pi*d), where d is the separation and mu_0 = 4*pi*10^-7 T m/A. This comes from wire 1 making a field B = mu_0*I1/(2*pi*d) at wire 2, and wire 2 feeling F = I2 * L * B. Note the force is per METRE of length, not a single total force, unless a length is given.

Is the force the same on both wires even if the currents are different?

Yes. By Newton's third law the two wires feel equal and opposite forces, even when I1 does not equal I2. The formula is symmetric in I1 and I2, so F/L is identical on each wire. A wire with double the current does NOT feel double the force.

How do you make one wire float (equilibrium) below or above another?

For a wire to hang in equilibrium, the magnetic force per unit length must exactly cancel its weight per unit length: mu_0*I1*I2/(2*pi*d) = lambda*g, where lambda is mass per unit length. If the upper wire attracts the lower one upward, the lower wire lifts off when magnetic force exceeds lambda*g.

How do I find the force on the MIDDLE wire of three parallel wires?

Compute the force per unit length from EACH outer wire separately using mu_0*I^2/(2*pi*d), then add them as VECTORS (not simple numbers). If the two forces are perpendicular, the resultant is sqrt(2) times one force. If they are opposite (equal currents same side), they cancel.

⚠️ The NEET trap
Currents in the same direction repel (like magnet poles), so parallel wires push apart.
Currents in the SAME direction ATTRACT; only OPPOSITE (anti-parallel) currents repel. This is opposite to the 'like poles repel' magnet rule.
🧠 NTA loves flipping attract/repel. Say it aloud: 'Same way come together.'

Real NEET questions

2017

An arrangement of three parallel straight wires placed perpendicular to the plane of paper carrying the same current I along the same direction is shown, with the middle wire B at distance d from each of the other two, and lines BA and BC at 90 degrees. The magnitude of the force per unit length on the middle wire B is:

A · mu_0 I^2 / (2 pi d)
B · sqrt(2) mu_0 I^2 / (pi d)
C · sqrt(3) mu_0 I^2 / (2 pi d)
D · sqrt(2) mu_0 I^2 / (2 pi d)
Solution: Step 1: Each outer wire is at distance d from B and carries the same current I in the same direction, so each attracts B. Step 2: Force per unit length from each = mu_0*I^2/(2*pi*d). Step 3: The two forces BA and BC are mutually perpendicular (90 degrees between them). Step 4: Resultant = sqrt(F^2 + F^2) = sqrt(2)*F = sqrt(2)*mu_0*I^2/(2*pi*d). Answer: D.
2026

Two infinitely long parallel wires A and B carry currents I and 2I in the same direction. Wire A (mass per unit length lambda) lies on an insulated floor; wire B is fixed at height h above it. The minimum h so that wire A does NOT rise from the floor is:

A · mu_0 I^2 / (2 pi lambda g)
B · mu_0 I^2 / (pi lambda g)
C · 2 mu_0 I^2 / (pi lambda g)
D · 4 mu_0 I^2 / (pi lambda g)
Solution: Step 1: Same-direction currents attract, so B pulls A upward. Step 2: Upward magnetic force per unit length = mu_0*(I)(2I)/(2*pi*h) = mu_0*I^2/(pi*h). Step 3: Wire A just begins to lift when this equals its weight per unit length: mu_0*I^2/(pi*h) = lambda*g. Step 4: Solve for h = mu_0*I^2/(pi*lambda*g). Smaller h gives a bigger pull, so this is the minimum height. Answer: B.

Solved Moving Charges And Magnetism NEET PYQs

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Frequently asked

Do parallel currents attract or repel?

Currents in the same direction attract; currents in opposite directions repel. This is the opposite of the 'like poles repel' rule for magnets, which is why it is a common trap.

What is the formula for force between two parallel wires?

Force per unit length F/L = mu_0 * I1 * I2 / (2 * pi * d), where d is the distance between the wires and mu_0 = 4*pi*10^-7 T m/A.

How is one ampere defined using this force?

One ampere is the steady current in each of two long parallel wires 1 metre apart that produces a force of 2*10^-7 N per metre of length on each wire. Putting I1 = I2 = 1 A and d = 1 m into the formula gives exactly 2*10^-7 N/m.

Are the forces on the two wires equal if the currents differ?

Yes. By Newton's third law and the symmetric formula, both wires feel the same magnitude of force per unit length, even if one carries more current.

When is a parallel wire in equilibrium?

When the magnetic force per unit length balances gravity: mu_0*I1*I2/(2*pi*d) = lambda*g, where lambda is the wire's mass per unit length.