NEET 2026 (Phase 1) · PhysicsAmpere thick wirePrevious Year Question
The figure shows a long straight solid wire of circular cross-section of radius 'a' carrying steady current I. The current I is uniformly distributed across its cross-section. The plot which correctly represents the variation of magnetic field (B) with distance (r) from the axis of the conductor is:
Answer: (A) B rises linearly for r < a, then falls as 1/r for r > a. Answer: (A) Solution: Apply Ampère's law.
- A.B rises linearly for r < a, then falls as 1/r for r > a✓
- B.B falls as 1/r for r < a, then rises linearly for r > a
- C.B is constant for r < a, then falls as 1/r for r > a
- D.B rises linearly for r < a, then stays constant for r > a
Correct Answer
(A) B rises linearly for r < a, then falls as 1/r for r > a
Solution & Explanation
Answer: (A) Solution: Apply Ampère's law. Inside (r < a) the enclosed current ∝ r²: B(2πr) = μ₀ I (r²/a²) → B = μ₀ I r /(2π a²), so B ∝ r (linear rise from 0). Outside (r > a) full current is enclosed: B(2πr) = μ₀ I → B = μ₀ I /(2π r), so B ∝ 1/r (decay). B is maximum at r = a. The correct plot rises linearly to r = a, then decays as 1/r.
🎯
43,000+ questions in हिंदी & English · one-tap toggle
Open →📚Practice all NEET Physics PYQs chapter-wise →💡Don't get it? Learn the Moving Charges And Magnetism concepts →Practice NEET 2026 (Phase 1) Physics — free, with instant solutions
43,000+ NEET questions solved step-by-step, chapter-wise, in the MedicNEET app.
More NEET PYQ solutions
PhysFrom Ampère's circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is:PhysA thick current cable of radius R carries current 'I' uniformly distributed across its cross-section. The variation of magnetic field B(r) due to the cable with the distance 'r' from the axis of the cable is represented by:PhysA cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the magnetic field B with the distance d from the centre of the conductor is correctly represented by the figure:PhysA long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B′ at radial distances a/2 and 2a respectively, from the axis of the wire, is:PhysAn electric dipole with dipole moment 5 × 10⁻⁶ C·m is aligned with the direction of a uniform electric field of magnitude 4 × 10⁵ N/C. The dipole is then rotated through an angle of 60° with respect to the electric field. The change in the potential energy of the dipole is:PhysTwo identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (radii negligible vs separation, treated as point charges) is best given as:PhysThe plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K1 and K2 with thickness 3d/8 and d/2 respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K1/K2 = 1.25, the value of K1 is:PhysA sphere of radius R is carved out of a uniform solid sphere of radius 2R (internally tangent, so its centre is at distance R from the big centre). The ratio of the moment of inertia of the small sphere to that of the remaining part, both about the Y-axis through the big sphere's centre, is: