Q.5 Physics ⭐ 360° Option Elimination Active

NEET 2026 Physics Question 5 Solution

Topic: Electromagnetic Induction · Official NTA Booklet Code 50

Question

Consider a long solenoid of length l and radius r. If n is the number of turns per unit length and μ₀ is the permeability of free space, the inductance of the solenoid is:

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Verified Against NTA Final Answer Key

Option (1) : L = mu_0 * pi * n^2 * r^2 * l

📚 Electromagnetic Induction Easy

📋 Given Parameters & Problem Setup

Length of Solenoid l meters

Radius of Solenoid r meters (cross-section area A = pi * r^2)

Turns per unit length n turns/m (total turns N = n * l)

Permeability mu_0 (free space)

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The 30-Second Exam Hall Shortcut: ⚡ Dimensional & Turn-Density Check

⏱️ Instant recall saves full 1 minute.

Self-inductance must depend quadratically on turn density n^2 and volume

1 Recall general formula: L = mu_0 * N^2 * A / l.

2 Express in turns per unit length n = N / l ==> N = n * l.

3 Substitute: L = mu_0 * (n * l)^2 * (pi * r^2) / l = mu_0 * pi * n^2 * r^2 * l.

4 Notice: Volume is (pi * r^2 * l). So L = mu_0 * n^2 * (Volume). Solved in 10 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Self-inductance represents magnetic energy per unit current squared: U = (1/2)*L*I^2 = (B^2 / 2mu_0) * Volume. Since B = mu_0 * n * I, U = (1/2) * mu_0 * n^2 * I^2 * (pi * r^2 * l), proving L = mu_0 * pi * n^2 * r^2 * l.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Calculate Magnetic Field Inside the Solenoid

For a long solenoid with n turns per unit length carrying current I, the uniform interior magnetic field is:

B = \mu_0 n I

• End effects are neglected for a long solenoid (l >> r).
• Field lines are axial and uniform throughout the interior cross-section.

Step 2

Calculate Magnetic Flux Linked with Total Turns

Cross-sectional area of one turn is A = pi * r^2. Total number of turns in length l is N = n * l. Total magnetic flux linkage Phi is:

\Phi = N \cdot (B A) = (n l) \cdot (\mu_0 n I) \cdot (\pi r^2) = \mu_0 \pi n^2 r^2 l I

• Each turn links flux Phi_1 = B * A = mu_0 * n * I * pi * r^2.
• Multiplying by total turns N = n * l gives total flux linkage.

Step 3

Extract Self-Inductance L

By definition of self-inductance L = Phi / I:

L = \frac{\Phi}{I} = \mu_0 \pi n^2 r^2 l

• Current I cancels completely, leaving self-inductance as a purely geometric property.
• Matches Option (1) 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)

✓ 100% Correct

Verdict & Scientific Reason:

Exact canonical formula for self-inductance of a circular solenoid.

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Why this is airtight: Direct derivation from Faraday flux linkage and Ampere's law.

Note: Matches Official NTA Final Answer Key.

2 Option (2)

✗ Incorrect

Verdict & Scientific Reason:

Missing factor of pi in cross-sectional area.

⚠️

Student Fallacy (Why candidates pick this trap): Mistakenly assumes cross-sectional area is r^2 instead of pi * r^2.

Note: For circular cross-section, area is always pi * r^2.

3 Option (3)

✗ Incorrect

Verdict & Scientific Reason:

Confuses solenoid field with long straight wire field factor (mu_0 / 2pi).

⚠️

Student Fallacy (Why candidates pick this trap): Inserts 2pi in denominator from Biot-Savart straight conductor formula B = mu_0 I / (2pi r).

Note: Solenoid field has NO 2pi in denominator: B = mu_0 * n * I.

4 Option (4)

✗ Incorrect

Verdict & Scientific Reason:

Spurious factor of 2.

⚠️

Student Fallacy (Why candidates pick this trap): Erroneously introduces a factor of 2, perhaps confusing diameter with radius.

Note: Derived directly with radius r; no factor of 2 exists.

⚠️

NTA Trap Alert: Biot-Savart Constant Confusion

Students often mix up the formula constants between straight wires (mu_0 / 2pi), circular loops (mu_0 / 2r), and solenoids (mu_0 * n).

🛡️ How to avoid the -1 mark: Anchor solenoid field: B = mu_0 * n * I. There is never a pi in the denominator for an ideal solenoid.

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Core NEET Syllabus Standard

NEET UG Physics Core Standards · Electromagnetic Induction Standard Benchmark

Core Scientific Principle Tested:

"L = mu_0 * n^2 * A * l = mu_0 * n^2 * V, where V is the interior volume of the solenoid."

Syllabus Focus: Self-Inductance of a Long Solenoid

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2026 Physics Question 5 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Dimensional & Turn-Density Check

Step-by-Step Rigorous Workflow & Mathematical Proof

Calculate Magnetic Field Inside the Solenoid

Calculate Magnetic Flux Linked with Total Turns

Extract Self-Inductance L

360° Option Elimination Engine Key Feature

NTA Trap Alert: Biot-Savart Constant Confusion

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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