Q.15 Physics ⭐ 360° Option Elimination Active

NEET 2026 Physics Question 15 Solution

Topic: Magnetic Effects of Current · Official NTA Booklet Code 50

Question

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.]

✓

Verified Against NTA Final Answer Key

Option (2) : h_min = mu_0 * I^2 / (pi * lambda * g)

📚 Magnetic Effects of Current Medium

📋 Given Parameters & Problem Setup

Wire A Current I (resting on insulated floor)

Wire B Current 2I (fixed at height h directly above wire A)

Current Directions Same direction (parallel currents)

Mass per unit length of Wire A lambda kg/m

Condition Wire A does not rise from the floor (Normal force N >= 0)

⚡

The 30-Second Exam Hall Shortcut: ⚡ Ampere Force Balance Shortcut

⏱️ Avoids vector integration.

Equate upward magnetic attraction to downward gravitational force

1 Parallel currents in the SAME direction ATTRACT each other.

2 Wire B pulls wire A upwards with magnetic force per unit length: F_m / L = mu_0 * I_A * I_B / (2pi * h).

3 Substitute currents: F_m / L = mu_0 * (I) * (2I) / (2pi * h) = mu_0 * I^2 / (pi * h).

4 Downward gravity per unit length: F_g / L = lambda * g.

5 For wire A to stay on the floor: F_m / L <= lambda * g ==> mu_0 * I^2 / (pi * h) <= lambda * g.

6 Rearrange for h: h >= mu_0 * I^2 / (pi * lambda * g). Solved in 25 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Ampere's Force Law gives F/L = mu_0*I_1*I_2/(2pi*d). The factor of 2 in wire B's current (2I) exactly cancels the factor of 2 in the 2pi denominator, leaving pi in the denominator.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Determine the Nature and Magnitude of the Magnetic Force

According to Ampere's force law, two parallel conductors carrying currents in the same direction attract each other. The magnetic force per unit length exerted on wire A by fixed wire B is:

\frac{F_m}{L} = \frac{\mu_0 I_A I_B}{2\pi h} = \frac{\mu_0 (I)(2I)}{2\pi h} = \frac{\mu_0 I^2}{\pi h}

• Because wire B is situated vertically above wire A, this attractive magnetic force acts vertically UPWARDS on wire A.

Step 2

Set Up Vertical Force Equilibrium on Wire A

Two vertical forces act on wire A resting on the insulated floor:

\sum F_y = N + \frac{F_m}{L} - \lambda g = 0 \implies N = \lambda g - \frac{\mu_0 I^2}{\pi h}

• N is the normal reaction from the floor per unit length.
• Downwards gravitational force per unit length is lambda * g.

Step 3

Apply the Condition for Contact (Does Not Rise)

For wire A not to rise from the floor, normal reaction must be non-negative (N >= 0):

N \ge 0 \implies \lambda g \ge \frac{\mu_0 I^2}{\pi h} \implies h \ge \frac{\mu_0 I^2}{\pi \lambda g}

• The minimum permissible height is h_{min} = \frac{\mu_0 I^2}{\pi \lambda g}.
• Matches Option (2) exactly.

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 Option (1)

✗ Incorrect

Verdict & Scientific Reason:

Omitted factor of 2 in current 2I.

⚠️

Student Fallacy (Why candidates pick this trap): Assumes both wires carry current I, leaving a factor of 2 in the denominator.

Note: Wire B carries 2I, which cancels the 2 in 2pi.

2 Option (2)

✓ 100% Correct

Verdict & Scientific Reason:

100% scientifically exact; confirmed by Official NTA Final Answer Key.

💡

Why this is airtight: Correctly applies Ampere attraction and normal contact equilibrium.

Note: Exact physics.

3 Option (3)

✗ Incorrect

Verdict & Scientific Reason:

Extra factor of 2.

⚠️

Student Fallacy (Why candidates pick this trap): Forgot to divide by 2 from the 2pi denominator of the straight conductor field.

Note: Formula is mu_0*I_1*I_2 / (2pi*h), not mu_0*I_1*I_2 / (pi*h).

4 Option (4)

✗ Incorrect

Verdict & Scientific Reason:

Squared current multiplier.

⚠️

Student Fallacy (Why candidates pick this trap): Squares the current 2I to get 4I^2.

Note: Current product is I * 2I = 2I^2, not (2I)^2 = 4I^2.

⚠️

NTA Trap Alert: Current Product vs Current Squared Confusion

Students often calculate (2I)^2 = 4I^2 instead of I_1 * I_2 = I * 2I = 2I^2, leading to Option (4).

🛡️ How to avoid the -1 mark: Always write the distinct wire currents separately: I_1 = I and I_2 = 2I ==> I_1 * I_2 = 2I^2.

📖

Core NEET Syllabus Standard

NEET UG Physics Core Standards · Moving Charges and Magnetism Standard Benchmark

Core Scientific Principle Tested:

"Magnetic force per unit length between two parallel straight currents: F/L = mu_0 * I_1 * I_2 / (2pi * d). Parallel currents attract; antiparallel repel."

Syllabus Focus: Force Between Two Parallel Current-Carrying Conductors

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2026 Physics Question 15 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Ampere Force Balance Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Determine the Nature and Magnitude of the Magnetic Force

Set Up Vertical Force Equilibrium on Wire A

Apply the Condition for Contact (Does Not Rise)

360° Option Elimination Engine Key Feature

NTA Trap Alert: Current Product vs Current Squared Confusion

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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