Q.3 Physics ⭐ 360° Option Elimination Active

NEET 2021 Physics Question 3 Solution

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

Question

An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of 105 m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant.

✓

Verified Against NTA Final Answer Key

Option (4) : 8 × 10^-20 N

📚 Moving Charges and Magnetism Easy

📋 Given Parameters & Problem Setup

Current in Wire (I) 5 A

Electron Speed (v) 10^5 m/s

Perpendicular Distance (r) 20 cm = 0.2 m

Elementary Charge (e) 1.6 × 10^-19 C

⚡

The 30-Second Exam Hall Shortcut: ⚡ 2-Step Lorentz Force: B = μ0 I / (2π r), F = q v B

⏱️ Save ~20 seconds

$B = \frac{2 \times 10^{-7} \times 5}{0.2} = 5 \times 10^{-6}\text{ T}$ . Force $F = e v B = (1.6 \times 10^{-19}) \times (10^5) \times (5 \times 10^{-6}) = 8 \times 10^{-20}\text{ N}$ ! Instant 15-second calculation matching Option (4)!

1 Compute B from wire:

B = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi \times 10^{-7})(5)}{2\pi (0.2)} = 5 \times 10^{-6}\text{ T}

2 Motion is parallel to wire, while B is perpendicular to the plane (

\theta = 90^\circ

3 Compute magnetic force:

F = q v B \sin 90^\circ = (1.6 \times 10^{-19})(10^5)(5 \times 10^{-6})

F = 8.0 \times 10^{-20}\text{ N}

5 Match Option (4).

🛡️

Why You Can 100% Trust This Shortcut: Canonical Biot-Savart law for infinite linear current and Lorentz force magnitude.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Calculating the Magnetic Field Produced by the Long Conductor

Using Ampere's circuital law / Biot-Savart law, the magnetic field at a perpendicular distance $r = 20\text{ cm} = 0.2\text{ m}$ from a long wire carrying current $I = 5\text{ A}$ is:

B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.2} = \frac{2 \times 10^{-7} \times 5}{0.2} = 5 \times 10^{-6}\text{ T}

Step 2

Determining the Angle Between Velocity and Magnetic Field

By the right-hand thumb rule, for current flowing from P to Q (towards the right), the magnetic field above the wire points out of the page ( $\hat{k}$ ). The electron moves parallel to the wire ( $\hat{i}$ ). Therefore, $\vec{v} \perp \vec{B}$ , so $\theta = 90^\circ$ and $\sin 90^\circ = 1$ .

Step 3

Calculating the Magnitude of the Magnetic Force

The magnitude of the magnetic Lorentz force on the electron ( $q = 1.6 \times 10^{-19}\text{ C}$ ) is:

F = |q| v B \sin 90^\circ = (1.6 \times 10^{-19}\text{ C}) \times (10^5\text{ m/s}) \times (5 \times 10^{-6}\text{ T}) = 8 \times 10^{-20}\text{ N}

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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:

Result of an arithmetic factor of 2 error (using $\mu_0 / 4\pi$ instead of $\mu_0 / 2\pi$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Using the semi-infinite wire formula

B = \mu_0 I / 4\pi r

2 2

✗ Incorrect

Verdict & Scientific Reason:

Fails to cancel $\pi$ between $\mu_0 = 4\pi \times 10^{-7}$ and the denominator $2\pi$ .

⚠️

Student Fallacy (Why candidates pick this trap): Forgetting that

\mu_0 / 2\pi = 2 \times 10^{-7}

without a leftover

\pi

3 3

✗ Incorrect

Verdict & Scientific Reason:

Combines both the $\pi$ cancellation error and the semi-infinite wire error.

⚠️

Student Fallacy (Why candidates pick this trap): Double mathematical substitution error.

4 4

✓ 100% Correct

Verdict & Scientific Reason:

Exact numerical result of the Lorentz force calculation.

💡

Why this is airtight: None; rigorous electrodynamics calculation.

⚠️

NTA Trap Alert: Pi Retention and Distance Unit Conversion Trap

Students frequently forget to convert $20\text{ cm}$ into $0.2\text{ m}$ , leading to power of 10 errors, or they fail to cancel $\pi$ from the denominator of $\mu_0 / 2\pi r$ .

🛡️ How to avoid the -1 mark: Use

\frac{\mu_0}{2\pi} = 2 \times 10^{-7}\text{ T}\cdot\text{m/A}

directly; convert

20\text{ cm} = 0.2\text{ m}

📖

Core NEET Syllabus Standard

NCERT Physics · Moving Charges and Magnetism Standard Benchmark

Core Scientific Principle Tested:

"The magnetic force on a charge $q$ moving with velocity $\vec{v}$ in field $\vec{B}$ is $\vec{F} = q(\vec{v} \times \vec{B})$ , where $B = \frac{\mu_0 I}{2\pi r}$ for an infinite straight wire."

Syllabus Focus: Magnetic Force on a Moving Charge and Magnetic Field of Straight Conductor

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Physics Question 3 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ 2-Step Lorentz Force: B = μ0 I / (2π r), F = q v B

Step-by-Step Rigorous Workflow & Mathematical Proof

Calculating the Magnetic Field Produced by the Long Conductor

Determining the Angle Between Velocity and Magnetic Field

Calculating the Magnitude of the Magnetic Force

360° Option Elimination Engine Key Feature

NTA Trap Alert: Pi Retention and Distance Unit Conversion Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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