Q.32 Physics ⭐ 360° Option Elimination Active

NEET 2021 Physics Question 32 Solution

Topic: Moving Charges and Magnetism · Official NTA Booklet Code M1

Question

A thick current carrying cable of radius 'R' carries current 'I' uniformly distributed across its cross-section. The variation of magnetic field B(r) due to the cable with the distance 'r' from the axis of the cable is represented by:

✓

Verified Against NTA Final Answer Key

Option (3) : Graph (3): Linear rise from origin inside (r <= R), hyperbolic decay outside (r >= R)

📚 Moving Charges and Magnetism Easy

📋 Given Parameters & Problem Setup

Cable Radius R

Current Distribution Uniformly distributed current I across cross-section

⚡

The 30-Second Exam Hall Shortcut: ⚡ Ampere's Law Signature: B ∝ r (Inside) and B ∝ 1/r (Outside)

⏱️ Save ~20 seconds

For uniform current density inside a thick cable: $B_{\text{in}} \propto r$ (straight line passing through the origin up to $r = R$ ). Outside the cable, total current is constant, so $B_{\text{out}} \propto 1/r$ (hyperbolic decay). Only Graph (3) displays this linear-increase followed by hyperbolic-decrease profile! Match Option (3)!

1 Inside (

r \le R

): Enclosed current

I_{\text{enc}} = I (r^2 / R^2)

. By Ampere's law,

B(2\pi r) = \mu_0 I (r^2 / R^2) \implies B \propto r

(straight line).

2 Outside (

r \ge R

): Enclosed current is total

I

. By Ampere's law,

B(2\pi r) = \mu_0 I \implies B \propto \frac{1}{r}

(hyperbolic curve).

3 At surface (

r = R

): Magnetic field attains its maximum value

B_{\text{max}} = \frac{\mu_0 I}{2\pi R}

4 Match Graph (3).

🛡️

Why You Can 100% Trust This Shortcut: Canonical Ampere's circuital law evaluation for solid cylindrical conductors.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Applying Ampere's Circuital Law Inside the Cable (r <= R)

Consider an Amperian loop of radius $r \le R$ concentric with the cable axis. Because current $I$ is uniformly distributed over the cross-sectional area $\pi R^2$ , the current enclosed by the loop is:

I_{\text{enc}} = J \cdot (\pi r^2) = \left(\frac{I}{\pi R^2}\right) (\pi r^2) = I \frac{r^2}{R^2}

Step 2

Finding the Magnetic Field Inside

By Ampere's circuital law $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$ :

B_{\text{in}} (2\pi r) = \mu_0 I \left(\frac{r^2}{R^2}\right) \implies B_{\text{in}} = \frac{\mu_0 I r}{2\pi R^2} \implies \mathbf{B_{\text{in}} \propto r}

Step 3

Finding the Magnetic Field Outside the Cable (r >= R)

For an Amperian loop of radius $r \ge R$ , the entire current $I$ is enclosed:

B_{\text{out}} (2\pi r) = \mu_0 I \implies B_{\text{out}} = \frac{\mu_0 I}{2\pi r} \implies \mathbf{B_{\text{out}} \propto \frac{1}{r}}

Step 4

Identifying the Correct Graphical Representation

The field starts at 0 at the center ( $r = 0$ ), rises linearly with distance $r$ up to the surface $r = R$ , where it reaches a peak value, and then decreases asymptotically as $1/r$ for $r > R$ . This corresponds uniquely to Graph (3).

⭐

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:

Falsely shows field increasing indefinitely with quadratic curvature.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing enclosed area (

pi r^2

) with field variation without dividing by circumference (

2pi r

2 2

✗ Incorrect

Verdict & Scientific Reason:

Shows $B propto r$ everywhere, ignoring that outside the cable the enclosed current no longer increases.

⚠️

Student Fallacy (Why candidates pick this trap): Extrapolating the internal field equation outside the conductor boundary.

3 3

✓ 100% Correct

Verdict & Scientific Reason:

Exact textbook graph showing $B \propto r$ for $r \le R$ and $B \propto 1/r$ for $r \ge R$ .

💡

Why this is airtight: None; canonical NCERT Ampere's law curve.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Fails to show linear behavior inside the cable.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing magnetic field profile with electrostatic potential curves.

⚠️

NTA Trap Alert: Solid vs. Hollow Cylinder Field Trap

For a hollow cylindrical pipe, $B = 0$ inside and $B \propto 1/r$ outside. For a SOLID cable carrying uniform current, $B$ is NOT zero inside—it increases linearly with $r$ ( $B \propto r$ ).

🛡️ How to avoid the -1 mark: Solid cable

\implies B_{\text{in}} \propto r

. Hollow pipe

\implies B_{\text{in}} = 0

📖

Core NEET Syllabus Standard

NCERT Physics · Moving Charges and Magnetism Standard Benchmark

Core Scientific Principle Tested:

"Inside a uniform wire, $B = \frac{\mu_0 I r}{2\pi R^2} \propto r$ ; outside, $B = \frac{\mu_0 I}{2\pi r} \propto \frac{1}{r}$ ."

Syllabus Focus: Magnetic Field Due to a Cylindrical Wire (Ampere's Circuital Law)

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Physics Question 32 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Ampere's Law Signature: B ∝ r (Inside) and B ∝ 1/r (Outside)

Step-by-Step Rigorous Workflow & Mathematical Proof

Applying Ampere's Circuital Law Inside the Cable (r <= R)

Finding the Magnetic Field Inside

Finding the Magnetic Field Outside the Cable (r >= R)

Identifying the Correct Graphical Representation

360° Option Elimination Engine Key Feature

NTA Trap Alert: Solid vs. Hollow Cylinder Field Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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