Q.44 Physics ⭐ 360° Option Elimination Active

NEET 2023 Physics Question 44 Solution

Topic: Moving Charges and Magnetism · Official NTA Booklet Code G5

Question

A wire carrying a current I along the positive x-axis has length L. It is kept in a magnetic field B = (2î + 3ĵ − 4k̂) T. The magnitude of the magnetic force acting on the wire is:

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Verified Against NTA Final Answer Key

Option (4) : 5 IL

📚 Moving Charges and Magnetism Easy

📋 Given Parameters & Problem Setup

Current element vector (L⃗) L î (along positive x-axis)

Magnetic field vector (B⃗) (2î + 3ĵ - 4k̂) T

Formula for magnetic force F⃗ = I (L⃗ × B⃗)

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The 30-Second Exam Hall Shortcut: ⚡ Vector Cross Product & 3-4-5 Triangle Hack

⏱️ Save ~40 seconds

î × î = 0, î × ĵ = k̂, î × k̂ = -ĵ. Thus F⃗ = IL [3k̂ - 4(-ĵ)] = IL (4ĵ + 3k̂). Magnitude = IL √(4² + 3²) = 5 IL in 15 seconds!

1 Express wire vector in Cartesian coordinates: L⃗ = L î.

2 Take vector cross product with B⃗ = 2î + 3ĵ - 4k̂:

3 - î × 2î = 0 (parallel component exerts zero force).

4 - î × 3ĵ = 3k̂.

5 - î × (-4k̂) = -4(-ĵ) = +4ĵ.

6 Net force vector: F⃗ = I L (4ĵ + 3k̂).

7 Calculate magnitude: |F⃗| = I L √(4² + 3²) = I L √(16 + 9) = I L √25 = 5 I L.

8 Directly choose Option (4) 5 IL!

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Why You Can 100% Trust This Shortcut: Magnetostatics · Lorentz Force on Conductors (Magnetic force on a straight conductor is F⃗ = I L⃗ × B⃗, perpendicular to both current and field).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Lorentz Force Law on a Current-Carrying Conductor

A straight wire of length vector L⃗ carrying a steady current I placed in a uniform magnetic field B⃗ experiences a total magnetic force given by the vector cross product:

\vec{F} = I (\vec{L} \times \vec{B})

Step 2

Definition of Vectors in Unit Vector Notation

The wire carries current I along the positive x-axis, so its length displacement vector is directed entirely along î:

\vec{L} = L\,\hat{i}, \qquad \vec{B} = 2\,\hat{i} + 3\,\hat{j} - 4\,\hat{k}

Step 3

Evaluation of the Vector Cross Product

Using the cyclic right-handed rules for unit vectors (î × î = 0, î × ĵ = k̂, î × k̂ = -ĵ):

\begin{aligned} \vec{L} \times \vec{B} &= (L\,\hat{i}) \times (2\,\hat{i} + 3\,\hat{j} - 4\,\hat{k}) \\[6pt] &= L \left[ 2(\hat{i} \times \hat{i}) + 3(\hat{i} \times \hat{j}) - 4(\hat{i} \times \hat{k}) \right] \\[6pt] &= L \left[ 0 + 3\,\hat{k} - 4(-\hat{j}) \right] = L (4\,\hat{j} + 3\,\hat{k}) \end{aligned}

Step 4

Magnitude Calculation

The magnitude of the resultant force vector is:

|\vec{F}| = I L \sqrt{4^2 + 3^2} = I L \sqrt{16 + 9} = I L \sqrt{25} = 5 I L

⭐

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:

Arbitrary radical choice.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate subtracted component squares incorrectly.

2 (2)

✗ Incorrect

Verdict & Scientific Reason:

Omitted the z-component (-4k̂) cross product.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate considered only the ĵ component of the field.

3 (3)

✗ Incorrect

Verdict & Scientific Reason:

Square root of 5 arithmetic error.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate added components without squaring properly.

4 (4)

✓ 100% Correct

Verdict & Scientific Reason:

Accurate force magnitude: √(4² + 3²) IL = 5 IL.

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Why this is airtight: None. Candidate executed cross product accurately.

⚠️

NTA Trap Alert: Including the Parallel B_x Component in Force Magnitude

The x-component of the magnetic field (2î) is PARALLEL to the wire (Lî). Since î × î = 0, parallel magnetic field produces ZERO magnetic force! Candidates who compute the total field magnitude |B| = √(2² + 3² + 4²) = √29 ≈ 5.38 IL commit this classic conceptual error.

🛡️ How to avoid the -1 mark: Remember: Only perpendicular components of magnetic field exert force on a current!

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Core NEET Syllabus Standard

National Eligibility cum Entrance Test (NEET UG) · Moving Charges and Magnetism Standard Benchmark

Core Scientific Principle Tested:

"Magnetic force is F⃗ = I(L⃗ × B⃗). Parallel field components produce zero force."

Syllabus Focus: Magnetic Force on a Current-Carrying Conductor

Preserved in Rationalized Curriculum Syllabus & Academic Standards ↗

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NEET 2023 Physics Question 44 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Vector Cross Product & 3-4-5 Triangle Hack

Step-by-Step Rigorous Workflow & Mathematical Proof

Lorentz Force Law on a Current-Carrying Conductor

Definition of Vectors in Unit Vector Notation

Evaluation of the Vector Cross Product

Magnitude Calculation

360° Option Elimination Engine Key Feature

NTA Trap Alert: Including the Parallel B_x Component in Force Magnitude

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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