Q.36 Physics ⭐ 360° Option Elimination Active

NEET 2021 Physics Question 36 Solution

Topic: Electromagnetic Induction · Official NTA Booklet Code M1

Question

Two conducting circular loops of radii R1 and R2 are placed in the same plane with their centres coinciding. If R1 >> R2, the mutual inductance M between them will be directly proportional to:

✓

Verified Against NTA Final Answer Key

Option (4) : R2^2 / R1

📚 Electromagnetic Induction Medium

📋 Given Parameters & Problem Setup

Outer Loop Radius R_1

Inner Loop Radius R_2

Geometric Condition R_1 >> R_2 (concentric and coplanar)

⚡

The 30-Second Exam Hall Shortcut: ⚡ Mutual Inductance Flux: M = Φ/I ∝ (Area_inner) × B_outer

⏱️ Save ~25 seconds

Magnetic field at center due to outer loop: $B_1 = \frac{\mu_0 I_1}{2 R_1} \propto \frac{1}{R_1}$ . Flux through small inner loop: $\Phi_{21} = B_1 A_2 \propto \left(\frac{1}{R_1}\right)(\pi R_2^2) \propto \frac{R_2^2}{R_1}$ . Thus, $M \propto \frac{R_2^2}{R_1}$ ! Match Option (4)!

1 Current

I_1

in large loop of radius

R_1

produces magnetic field at common center:

B_1 = \frac{\mu_0 I_1}{2 R_1}

2 Since

R_1 \gg R_2

, the field

B_1

is essentially uniform over the area of the smaller loop (

A_2 = \pi R_2^2

3 Magnetic flux linked with inner loop:

\Phi_{21} = B_1 A_2 = \left(\frac{\mu_0 I_1}{2 R_1}\right)(\pi R_2^2)

4 By definition, mutual inductance is

M = \frac{\Phi_{21}}{I_1} = \frac{\mu_0 \pi R_2^2}{2 R_1}

5 Therefore,

M \propto \frac{R_2^2}{R_1}

6 Match Option (4).

🛡️

Why You Can 100% Trust This Shortcut: Canonical Biot-Savart center field and mutual flux coupling geometry.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Calculating Magnetic Field Produced by the Outer Loop

Let a current $I_1$ pass through the outer circular loop of radius $R_1$ . By the Biot-Savart law, the magnetic field at the center of the loop is directed perpendicular to the plane of the loop:

B_1 = \frac{\mu_0 I_1}{2 R_1}

Step 2

Determining the Magnetic Flux Linked with the Inner Loop

Because $R_1 \gg R_2$ , the magnetic field $B_1$ can be treated as virtually uniform across the cross-sectional area of the inner loop $A_2 = \pi R_2^2$ . The flux $\Phi_{21}$ linked with loop 2 is:

\Phi_{21} = B_1 \cdot A_2 = \left(\frac{\mu_0 I_1}{2 R_1}\right) \left(\pi R_2^2\right) = \frac{\mu_0 \pi R_2^2}{2 R_1} I_1

Step 3

Extracting the Mutual Inductance M

Mutual inductance is defined as the flux linkage in coil 2 per unit current in coil 1:

M = \frac{\Phi_{21}}{I_1} = \frac{\mu_0 \pi R_2^2}{2 R_1} \implies M \propto \frac{R_2^2}{R_1}

⭐

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 1

✗ Incorrect

Verdict & Scientific Reason:

Dimensionless ratio with inverted power of radii.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming mutual inductance depends linearly on radius ratio.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Missing the quadratic area factor $R_2^2$ of the inner loop.

⚠️

Student Fallacy (Why candidates pick this trap): Forgetting that magnetic flux requires multiplying field by area (

pi R_2^2

3 3

✗ Incorrect

Verdict & Scientific Reason:

Inverts which radius is squared and which is in the denominator.

⚠️

Student Fallacy (Why candidates pick this trap): Squaring the outer radius instead of the inner loop area radius.

4 4

✓ 100% Correct

Verdict & Scientific Reason:

Exact proportionality: $M = \frac{\mu_0 \pi R_2^2}{2 R_1} \propto \frac{R_2^2}{R_1}$ .

💡

Why this is airtight: None; rigorous mutual inductance derivation.

⚠️

NTA Trap Alert: Radius Squaring Inversion Trap

Students who confuse the field-producing loop with the flux-collecting loop often calculate $M \propto R_1^2 / R_2$ (Option 3). The inner loop collects flux over its area ( $A_2 \propto R_2^2$ ), while the outer loop creates field ( $B_1 \propto 1/R_1$ ).

🛡️ How to avoid the -1 mark: Remember:

M \propto (\text{Area of small}) / (\text{Radius of large}) = R_2^2 / R_1

📖

Core NEET Syllabus Standard

NCERT Physics · Electromagnetic Induction Standard Benchmark

Core Scientific Principle Tested:

"Mutual inductance between two concentric coplanar loops with $R_1 \gg R_2$ is $M = \frac{\mu_0 \pi R_2^2}{2 R_1}$ ."

Syllabus Focus: Mutual Inductance

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Physics Question 36 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Mutual Inductance Flux: M = Φ/I ∝ (Area_inner) × B_outer

Step-by-Step Rigorous Workflow & Mathematical Proof

Calculating Magnetic Field Produced by the Outer Loop

Determining the Magnetic Flux Linked with the Inner Loop

Extracting the Mutual Inductance M

360° Option Elimination Engine Key Feature

NTA Trap Alert: Radius Squaring Inversion Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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