Q.47 Physics ⭐ 360° Option Elimination Active

NEET 2023 Physics Question 47 Solution

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

Question

A very long conducting wire is bent in a semi-circular shape from A to B as shown in figure. The magnetic field at point P for steady current configuration is given by:

✓

Verified Against NTA Final Answer Key

Option (4) : (μ₀ i / 4R) [1 - 2/π] pointed away from page

📚 Moving Charges and Magnetism Medium

📋 Given Parameters & Problem Setup

Current in conductor i

Radius of semicircular arc AB R (centered at observation point P)

Top straight segment Semi-infinite straight wire carrying current to the right

Bottom straight segment Semi-infinite straight wire carrying current to the left

⚡

The 30-Second Exam Hall Shortcut: ⚡ Biot-Savart Superposition & Direction Inspection

⏱️ Save ~60 seconds

Semicircular arc at center P: B_arc = μ₀i / (4R) ⊙ (out of page). Two semi-infinite straight wires: each is μ₀i / (4πR) ⊗ (into page) ⟹ B_straight = 2 × [μ₀i / (4πR)] = μ₀i / (2πR) ⊗. Net B = (μ₀i / 4R)[1 - 2/π] ⊙ (away from page) in 30 seconds!

1 Decompose the current geometry into three independent parts:

2 - Top semi-infinite straight wire ending at A.

3 - Semicircular arc AB of radius R centered at P.

4 - Bottom semi-infinite straight wire starting at B.

5 Arc field at center P: B_arc = (1/2) × (μ₀i / 2R) = μ₀i / 4R. By right-hand grip rule (current downwards along arc), thumb points OUT OF PAGE (⊙).

6 Top straight wire at distance R below its terminus: B₁ = μ₀i / (4πR) directed INTO PAGE (⊗).

7 Bottom straight wire at distance R above its terminus: B₂ = μ₀i / (4πR) directed INTO PAGE (⊗).

8 Total straight wire field: B_straight = B₁ + B₂ = μ₀i / (2πR) ⊗.

9 Subtract (since ⊙ and ⊗ oppose each other): B_net = B_arc - B_straight = (μ₀i / 4R) - (μ₀i / 2πR) = (μ₀i / 4R) [1 - 2/π].

10 Because 1 > 2/π (1 > 0.637), B_arc dominates ⟹ net direction is OUT OF PAGE (away from page).

11 Select Option (4)!

🛡️

Why You Can 100% Trust This Shortcut: Magnetostatics · Biot-Savart Law (Total magnetic field is vector sum of contributions from all current elements: B⃗ = ∮ dB⃗).

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Magnetic Field Due to the Semicircular Arc (B_arc)

By the Biot-Savart Law, a circular loop of radius R carrying current i produces B = μ₀i / 2R at its center. A semicircle subtends an angle of π radians at center P, producing exactly half this value:

B_{\text{arc}} = \frac{1}{2} \left( \frac{\mu_0 i}{2R} \right) = \frac{\mu_0 i}{4R} \quad (\odot \text{ directed out of the page})

Step 2

Magnetic Field Due to the Top Semi-Infinite Straight Wire (B₁)

The top wire extends from x = -∞ to x = A at a perpendicular distance R above point P. By the standard Biot-Savart formula for a semi-infinite straight wire at an endpoint perpendicular distance R:

B_1 = \frac{\mu_0 i}{4\pi R} \quad (\otimes \text{ directed into the page})

Step 3

Magnetic Field Due to the Bottom Semi-Infinite Straight Wire (B₂)

The bottom wire carries current i to the left from x = B to x = -∞ at a perpendicular distance R below point P. Applying the right-hand grip rule, its field at point P also points INTO the page:

B_2 = \frac{\mu_0 i}{4\pi R} \quad (\otimes \text{ directed into the page})

Step 4

Vector Superposition of Opposing Fields

The arc field points out of the page (⊙), while the sum of both straight wire fields points into the page (⊗). Subtracting the opposing magnitudes:

B_{\text{net}} = B_{\text{arc}} - (B_1 + B_2) = \frac{\mu_0 i}{4R} - \frac{2\mu_0 i}{4\pi R} = \frac{\mu_0 i}{4R} \left[ 1 - \frac{2}{\pi} \right]

Step 5

Direction of Resultant Field

Since π ≈ 3.1416, 2/π ≈ 0.637 < 1, the bracket [1 - 2/π] is strictly positive (+0.363). Therefore, the outward arc field dominates, and the net magnetic field points AWAY FROM (out of) the page.

\vec{B}_{\text{net}} = \frac{\mu_0 i}{4R} \left[ 1 - \frac{2}{\pi} \right] \quad (\text{pointed away from the page})

⭐

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:

Wrong direction: arc field is stronger than straight wire fields (1 > 2/π).

⚠️

Student Fallacy (Why candidates pick this trap): Candidate incorrectly assumed the straight wires dominate the arc.

2 (2)

✗ Incorrect

Verdict & Scientific Reason:

Omitted the straight wire contributions entirely and inverted direction.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate forgot the two semi-infinite wires.

3 (3)

✗ Incorrect

Verdict & Scientific Reason:

Represents only the arc field, ignoring the straight wires.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate calculated only the semicircular arc field.

4 (4)

✓ 100% Correct

Verdict & Scientific Reason:

Accurate magnitude and outward direction matching Biot-Savart Law and NTA final answer key.

💡

Why this is airtight: None. Candidate executed complete vector superposition.

⚠️

NTA Trap Alert: Directional Error: Not Knowing Whether 1 or 2/π is Greater

Since π ≈ 3.14, 2/π ≈ 0.64, which is LESS than 1. This means the arc field (pointing OUT) is larger than the straight wire fields (pointing IN). The net field must point OUT (away from the page). Choosing "into the page" (Option 1) is a common direction trap.

🛡️ How to avoid the -1 mark: Numerically compare: Arc factor is 1; Straight wires factor is 2/π ≈ 0.64. Since 1 > 0.64, Arc wins ⟹ Direction is OUT of page!

📖

Core NEET Syllabus Standard

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

Core Scientific Principle Tested:

"Magnetic field at center of an arc is B = (θ/2π)(μ₀i/2R); for a semi-infinite straight wire at endpoint, B = μ₀i/(4πR)."

Syllabus Focus: Biot-Savart Law and Applications to Wires and Arcs

Preserved in Rationalized Curriculum Syllabus & Academic Standards ↗

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NEET 2023 Physics Question 47 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Biot-Savart Superposition & Direction Inspection

Step-by-Step Rigorous Workflow & Mathematical Proof

Magnetic Field Due to the Semicircular Arc (B_arc)

Magnetic Field Due to the Top Semi-Infinite Straight Wire (B₁)

Magnetic Field Due to the Bottom Semi-Infinite Straight Wire (B₂)

Vector Superposition of Opposing Fields

Direction of Resultant Field

360° Option Elimination Engine Key Feature

NTA Trap Alert: Directional Error: Not Knowing Whether 1 or 2/π is Greater

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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