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 ↗

Ready to attempt under real exam pressure?

Test your speed and accuracy in the authentic 180-minute timed NTA simulator.

Launch NEET 2026 CBT Test →

← Previous (Q.10) All Physics Questions Grid Next (Q.12) →

Spotted a mistake in this solution? Report an error (support@mycompetitiveexams.com)

⚖️ Legal & Fair Dealing Notice: Sourced from publicly released NEET 2026 question papers and NEET answer keys for private study, research, and non-commercial educational practice under Fair Dealing (Section 52(1)(a) of the Indian Copyright Act, 1957). Not affiliated with or endorsed by NTA or NMC.

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?

🔗 Related Questions — Moving Charges and Magnetism

Crossed Electric and Magnetic Fields (Lorentz Force Condition)

Magnetic Moment Ratio for Circular Coils with Given Radii

Magnetic Field at the Center of a Circular Coil

Vector Lorentz Magnetic Force: F⃗ = I (L⃗ × B⃗)

Solve 4 NEET Questions a Day — One per Subject