B vs Distance Graph for a Solid Cylindrical Conductor
Physics · Moving Charges And Magnetism · NEET
For a solid cylindrical wire of radius R carrying uniform current, the magnetic field B rises in a straight line inside (B = μ₀Ir/2πR², so B ∝ r), reaches its MAXIMUM at the surface r = R, then falls off outside as B = μ₀I/2πr (so B ∝ 1/r). Memory hook: "Straight up to the skin, then curve down" — the peak is always at the surface, never at the centre (B = 0 at the axis).
B is zero at the axis, rises as a straight line (B ∝ r) up to the surface r = R where it peaks at B_max = μ₀I/2πR, then falls as a 1/r curve outside. The red line is the inside region; the blue curve is the outside region.
Your doubts, answered
Why is the magnetic field zero at the centre (axis) of the wire?
Ampere's law says B×(2πr) = μ₀ × (current enclosed by the loop). At the exact axis, r = 0, so the Amperian loop encloses zero current. With no enclosed current and zero loop radius, B = 0. As you move out, the loop starts to enclose current, so B grows from zero.
Why does B increase INSIDE but decrease OUTSIDE?
Inside (r < R), only part of the current is enclosed. Because current is spread evenly, the enclosed current grows as r² (area ∝ r²). So B = μ₀I·(r²/R²)/(2πr) = μ₀Ir/2πR² → B ∝ r (rises). Outside (r > R), the WHOLE current I is enclosed and stays fixed, so B = μ₀I/2πr → B ∝ 1/r (falls). The two effects are why the graph turns over at the surface.
Where exactly is B maximum?
At the surface, r = R. Put r = R into either formula and both give the same value B_max = μ₀I/2πR. The inside line and the outside curve meet here — this join point is the peak. Many students wrongly mark the centre as the maximum.
Is the field continuous at the boundary r = R?
Yes. Inside formula at r = R: μ₀IR/2πR² = μ₀I/2πR. Outside formula at r = R: μ₀I/2πR. Same value. So there is NO jump in B at the surface — the graph is a single connected curve with a sharp corner (a kink) at r = R, not a break.
How is this different from a hollow (thin) tube?
For a solid conductor, B rises linearly from 0 inside. For a hollow cylinder (current only on the outer skin), the enclosed current inside the hollow region is zero, so B = 0 everywhere inside, then jumps to μ₀I/2πR at the surface and falls as 1/r outside. NEET often mixes these two graphs as distractors.
⚠️ The NEET trap ✗ Picking the graph where B is maximum at the centre (r = 0) and falls outward the whole way. ✓ B is ZERO at the centre, rises linearly to a PEAK at the surface r = R, then falls as 1/r. The maximum sits at the surface, not the axis. 🧠 Solid wire = 'start at zero, climb to the skin, slide down'. Only the hollow tube stays flat at zero inside.
Real NEET questions
2019
A 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:
A · A (B constant inside, 1/d outside)
B · B (B falls inside, rises outside)
C · C (B ∝ d inside up to R, then 1/d outside) ✓
D · D (B rises inside, constant outside)
Solution: By Ampere's law for a uniform-current solid conductor. Inside (d < R): enclosed current ∝ d², so B(2πd) = μ₀I(d²/R²) → B = μ₀Id/2πR², i.e. B ∝ d (a straight line rising from 0). Outside (d > R): full current enclosed, B(2πd) = μ₀I → B = μ₀I/2πd, i.e. B ∝ 1/d (decays). B is maximum at d = R. The graph that rises linearly up to R then falls as 1/d is option C.
2022
From Ampere'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:
A · Uniform and constant for both regions
B · Linearly increasing up to the boundary, then linearly decreasing outside
C · Linearly increasing up to the boundary, then decreasing with 1/r dependence outside ✓
D · Linearly decreasing up to the boundary, then linearly increasing outside
Solution: Inside (r < R): B = μ₀Ir/2πR², so B increases LINEARLY with r. Outside (r > R): B = μ₀I/2πr, so B decreases as 1/r (NOT linearly). B peaks at the surface r = R. The correct description is 'linearly increasing up to the boundary, then 1/r decreasing outside' → option C. Trap option B wrongly says the outside falls linearly.
2016
A long straight wire of radius a carries a steady current I, 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:
A · 1/4
B · 1/2
C · 1 ✓
D · 4
Solution: Point 1 is INSIDE (r = a/2 < a): use B = μ₀Ir/2πa² = μ₀I(a/2)/2πa² = μ₀I/4πa. Point 2 is OUTSIDE (r = 2a > a): use B′ = μ₀I/2πr = μ₀I/2π(2a) = μ₀I/4πa. Both equal μ₀I/4πa, so ratio B : B′ = 1 → option C. Key skill: choose the INSIDE formula for a/2 and the OUTSIDE formula for 2a.
Solved Moving Charges And Magnetism NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula for B inside a solid cylindrical conductor?
B = μ₀Ir/2πR², valid for r < R, where R is the wire radius and r is the distance from the axis. This gives B ∝ r, a straight line rising from 0 at the centre to μ₀I/2πR at the surface.
What is the formula for B outside the conductor?
B = μ₀I/2πr, valid for r > R. This is the same as a thin straight wire, because outside the conductor the whole current acts as if concentrated on the axis. Here B ∝ 1/r, so it decreases as you go further away.
At what distance is the magnetic field the strongest?
At the surface, r = R. The maximum value is B_max = μ₀I/2πR. Both the inside and outside formulas give this same value at r = R, so the peak sits exactly on the boundary.
What does the B vs r graph look like overall?
Start at (0, 0), rise as a straight line up to the point (R, μ₀I/2πR), then curve downward following a 1/r hyperbola. It looks like a triangle-edge going up followed by a smooth falling tail, with a sharp corner at r = R.
Why do we assume uniform current distribution?
For a steady (DC) current in a normal conductor, charge spreads evenly over the cross-section, giving constant current density J = I/πR². This uniform J is what makes the enclosed current grow as r² inside and gives the neat linear B ∝ r result. High-frequency AC pushes current to the skin — a different case not tested at NEET level.
