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Q.34 Physics · Section A ⭐ 360° Option Elimination Active
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NEET 2024 Physics Question 34 Solution

Topic: Moving Charges and Magnetism · Official NTA Booklet Code T1
NEET 2024 Physics Question 34: Magnetic Field at the Center of a Circular Coil

Question

A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as 4π×10⁻⁷ SI units):

✓
Verified Against NTA Final Answer Key
Option (1) : 4.4 mT
📚 Moving Charges and Magnetism Easy
📋 Given Parameters & Problem Setup
Number of turns N = 100
Radius of coil R = 10 cm = 0.1 m
Current in coil I = 7 A
Permeability of free space μ₀ = 4π × 10⁻⁷ SI units (take π ≈ 22/7)
⚡

The 30-Second Exam Hall Shortcut: ⚡ 22/7 Fraction Cancellation Hack

⏱️ Save ~45 seconds

B = μ₀NI / 2R -> [4 × (22/7) × 10⁻⁷ × 100 × 7] / (2 × 0.1) -> 7 cancels with 7 -> 88 × 10⁻⁵ / 0.2 = 4.4 × 10⁻³ T = 4.4 mT!

1 Standard formula for center of circular coil: B = μ₀ N I / (2 R).
2 Write μ₀ = 4π × 10⁻⁷ with π = 22/7 so the given 7 A current cancels the denominator 7!
3 Numerator = 4 × (22/7) × 10⁻⁷ × 100 × 7 = 4 × 22 × 100 × 10⁻⁷ = 88 × 10⁻⁵.
4 Denominator = 2 × R = 2 × 0.1 = 0.2.
5 B = 88 × 10⁻⁵ / 0.2 = 440 × 10⁻⁵ = 4.4 × 10⁻³ T = 4.4 mT.
6 Option (1) in under 15 seconds!
🛡️
Why You Can 100% Trust This Shortcut: Biot-Savart Law for Circular Current Loops (B_center = μ₀ N I / (2R)).
📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Biot-Savart Formula for a Circular Coil

The magnitude of the magnetic field B at the center of a flat circular coil consisting of N closely wound turns carrying steady current I is given by the Biot-Savart law:

B=μ0NI2RB = \frac{\mu_0 N I}{2 R}
Step 2

Substitution of Given Parameters

Substitute N = 100, I = 7 A, R = 10 cm = 0.1 m, and μ₀ = 4π × 10⁻⁷ T m/A:

B=(4π×10−7)×100×72×0.1B = \frac{(4\pi \times 10^{-7}) \times 100 \times 7}{2 \times 0.1}
Step 3

Fractional Simplification using π = 22/7

Using π = 22/7 enables exact cancellation with the current of 7 A:

B=4×227×10−7×7000.2=4×22×10−50.2=88×10−50.2B = \frac{4 \times \frac{22}{7} \times 10^{-7} \times 700}{0.2} = \frac{4 \times 22 \times 10^{-5}}{0.2} = \frac{88 \times 10^{-5}}{0.2}
Step 4

Numerical Evaluation in Millitesla

Dividing 88 by 0.2 gives 440:

B=440×10−5 T=4.4×10−3 T=4.4 mTB = 440 \times 10^{-5}\,\text{T} = 4.4 \times 10^{-3}\,\text{T} = 4.4\,\text{mT}
⭐

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)
✓ 100% Correct
Verdict & Scientific Reason:

Exact evaluation: B = 4.4 × 10⁻³ T = 4.4 mT.

💡
Why this is airtight: None.
2 (2)
✗ Incorrect
Verdict & Scientific Reason:

Off by four orders of magnitude (44 T is an immense, laboratory-record magnetic field).

⚠️
Student Fallacy (Why candidates pick this trap): Unit conversion slip.
3 (3)
✗ Incorrect
Verdict & Scientific Reason:

Off by a factor of 10.

⚠️
Student Fallacy (Why candidates pick this trap): Misplacing the decimal point during 88 / 0.2 division.
4 (4)
✗ Incorrect
Verdict & Scientific Reason:

Confuses millitesla (mT) with tesla (T).

⚠️
Student Fallacy (Why candidates pick this trap): Ignoring the 10⁻³ factor.
⚠️

NTA Trap Alert: Unit Prefix Trap (mT vs T)

Options (1) and (4) differ solely by the prefix "m". Students frequently calculate 4.4 and overlook whether the unit is T or mT.

🛡️ How to avoid the -1 mark: A standard bench coil carrying 7 A produces a millitesla-range field (a few mT). 4.4 T requires superconducting MRI magnets!
📖

Core NEET Syllabus Standard

Physics Core Electromagnetism · Moving Charges and Magnetism Standard Benchmark
Core Scientific Principle Tested:

"The magnetic field at the center of a circular coil of N turns is B = μ₀NI / (2R)."

Syllabus Focus: Magnetic Field on the Axis and Center of a Circular Current Loop
✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗
7 A kills the 7 in π: 88 divided by 0.2 gives 4.4 mT!

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