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Physics Question 15 – NEET-UG 2026

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Two infinitely long parallel conducting wires A and B carry currents I and 2I, respectively, in the same direction. The wire A has uniform mass per unit length λ and lies on an insulated floor. The wire B is kept fixed at a height h above the floor. The minimum magnitude of h so that the wire A does not rise from the floor is : [g is the acceleration due to gravity and μ0 is the permeability of free space.]

Identify all forces acting on wire A.

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Ninja Strategy•Correct Force Calculation

Accurately calculate the magnetic force per unit length, remembering the product of currents (I)(2I) and the standard formula, then equate it to the gravitational force per unit length to find h.

Step 1: Identify Forces on Wire A✦ Active

Wire A experiences a downward gravitational force due to its mass and an upward attractive magnetic force due to wire B, as the currents are in the same direction. For wire A not to rise, the upward magnetic force must be balanced by the downward gravitational force.

Step 2: Calculate Forces per Unit Length○ Expand

The gravitational force per unit length on wire A is Fg/L=λg. The magnetic force per unit length between wire A (current IA=I) and wire B (current IB=2I) separated by a distance h is:

FmL=μ0IAIB2πh=μ0I(2I)2πh=μ02I22πh=μ0I2πh
💡 Teacher's Secret Hint

Remember that currents in the same direction result in an attractive force.

Step 3: Apply Equilibrium Condition and Solve for h○ Expand

For wire A not to rise, the upward magnetic force per unit length must be equal to the downward gravitational force per unit length:

FmL=FgL

Substituting the expressions from Step 2:

μ0I2πh=λg

Solving for h:

h=μ0I2πλg

This matches option (2).

💡 Teacher's Secret Hint

The 'minimum magnitude of h so that the wire A does not rise' implies the condition where the upward magnetic force just balances the gravitational force.

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