Q.11 Physics ⭐ 360° Option Elimination Active

NEET 2026 Physics Question 11 Solution

Topic: Magnetic Effects of Current · Official NTA Booklet Code 50

Question

A current I₀ flows through a metallic circular loop of radius r as shown in the figure. Resistance of the segment ABC is half that of ADC. Magnitude of magnetic field at the center O of the loop is:

✓

Verified Against NTA Final Answer Key

Option (1) : B_net = mu_0 * I_0 / (12 * r)

📚 Magnetic Effects of Current Medium

📋 Given Parameters & Problem Setup

Total Current Entering I_0 at node A and exiting at node C

Loop Radius r meters

Geometry Two semicircular branches ABC and ADC spanning angle pi radians

Resistance Relation R_ABC = (1/2) * R_ADC

Target Field Net magnetic field at center O

⚡

The 30-Second Exam Hall Shortcut: ⚡ Current Splitting & Net Current Shortcut

⏱️ Saves 2 minutes versus calculating two separate vector integrals.

Find differential current flowing through the two semicircles

1 Branches ABC and ADC are in parallel. Potential difference V is identical across both.

2 Since R_ABC = R_ADC / 2, current is inversely proportional to resistance: I_ABC = 2 * I_ADC.

3 Total current I_0 splits as: I_ABC = (2/3) * I_0, and I_ADC = (1/3) * I_0.

4 By right-hand rule, branch ABC produces field INTO the page (x), while ADC produces field OUT OF the page (.).

5 Net effective current = I_ABC - I_ADC = (2/3 - 1/3) * I_0 = (1/3) * I_0.

6 Magnetic field from a semicircle: B = (1/2) * [mu_0 * I_net / (2r)] = (1/4) * mu_0 * [(1/3)*I_0] / r = mu_0 * I_0 / (12r). Done in 25 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Biot-Savart Law states B = (theta / 2pi) * (mu_0 * I / 2r). For a semicircle (theta = pi), B = mu_0 * I / (4r). Since the two branches carry opposite circulating currents, their fields subtract vectorially: B_net = [mu_0 / (4r)] * (I_ABC - I_ADC).

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Calculate Current in Each Semicircular Branch

Semicircular segments ABC and ADC are connected in parallel between nodes A and C:

V = I_{ABC} R_{ABC} = I_{ADC} R_{ADC}

• Given R_{ABC} = \frac{1}{2} R_{ADC} \implies I_{ABC} = 2 I_{ADC}.
• Kirchhoff's Current Law at node A: I_{ABC} + I_{ADC} = I_0 \implies 2 I_{ADC} + I_{ADC} = 3 I_{ADC} = I_0.
• Therefore: I_{ADC} = \frac{1}{3} I_0, \quad I_{ABC} = \frac{2}{3} I_0.

Step 2

Determine Field Magnitudes and Directions at Center O

Use the Biot-Savart law for a circular arc spanning angle pi:

B_{\text{arc}} = \frac{\mu_0 I}{4\pi r} \cdot \theta = \frac{\mu_0 I}{4\pi r} (\pi) = \frac{\mu_0 I}{4r}

• Branch ABC (current flows A -> B -> C, clockwise): Magnetic field B_{ABC} points perpendicularly INTO the page (\otimes): B_{ABC} = \frac{\mu_0 (2/3 I_0)}{4r} = \frac{2\mu_0 I_0}{12r}.
• Branch ADC (current flows A -> D -> C, counter-clockwise): Magnetic field B_{ADC} points perpendicularly OUT OF the page (\odot): B_{ADC} = \frac{\mu_0 (1/3 I_0)}{4r} = \frac{\mu_0 I_0}{12r}.

Step 3

Superpose the Opposing Magnetic Fields

Subtract the smaller field from the larger field to obtain net magnitude:

B_{\text{net}} = B_{ABC} - B_{ADC} = \frac{2\mu_0 I_0}{12r} - \frac{\mu_0 I_0}{12r} = \frac{\mu_0 I_0}{12r}

• The net magnetic field points into the page with magnitude mu_0 * I_0 / (12r).
• Matches Option (1) exactly.

⭐

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 Option (1)

✓ 100% Correct

Verdict & Scientific Reason:

100% mathematically and physically exact; confirmed by Official NTA Final Answer Key.

💡

Why this is airtight: Correctly partitions parallel current in 2:1 ratio and subtracts semicircular contributions.

Note: Flawless execution.

2 Option (2)

✗ Incorrect

Verdict & Scientific Reason:

Uses full current I_0 in single semicircle.

⚠️

Student Fallacy (Why candidates pick this trap): Assumes the entire current I_0 flows through one branch while the other carries zero.

Note: Current must split across both parallel branches.

3 Option (3)

✗ Incorrect

Verdict & Scientific Reason:

Complete circle field without branch cancellation.

⚠️

Student Fallacy (Why candidates pick this trap): Uses full loop formula B = mu_0 * I / (2r) without dividing current or accounting for opposite directions.

Note: Two halves carry opposing currents.

4 Option (4)

✗ Incorrect

Verdict & Scientific Reason:

Straight wire formula confusion.

⚠️

Student Fallacy (Why candidates pick this trap): Applies Ampere's straight conductor formula instead of circular arc formula.

Note: Circular arcs have no pi in the denominator of their magnetic field.

⚠️

NTA Trap Alert: The Zero-Field False Cancellation Trap

Many candidates remember that "current entering and leaving a circular loop produces zero field at the center" — but that theorem ONLY holds when resistances are UNIFORM! When resistance is asymmetric (1:2), field does not cancel to zero.

🛡️ How to avoid the -1 mark: Check for symmetry: If R_1 = R_2, B = 0. If R_1 != R_2, currents are unequal and a non-zero net field remains!

📖

Core NEET Syllabus Standard

NEET UG Physics Core Standards · Moving Charges and Magnetism Standard Benchmark

Core Scientific Principle Tested:

"B = (mu_0 * I / 4pi * r) * theta. For parallel loop segments, net field depends on differential current Delta I = I_1 - I_2."

Syllabus Focus: Biot-Savart Law and Magnetic Field of Circular Arcs

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2026 Physics Question 11 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Current Splitting & Net Current Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Calculate Current in Each Semicircular Branch

Determine Field Magnitudes and Directions at Center O

Superpose the Opposing Magnetic Fields

360° Option Elimination Engine Key Feature

NTA Trap Alert: The Zero-Field False Cancellation Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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