The magnetic field due to a straight long conductor of uniform cross-section of radius a and carrying a steady current is represented by . In each graph B is the magnitude of the field and r is the distance from the axis of the conductor.

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(2) (2). Idea: apply Ampere's law ∮⃗B· d⃗l=μ₀I_enc to a circle of radius r centred on the axis. By symmetry B has the same magnitude all round it, so B(2π r)=μ₀I_enc. Inside the wire (r<a) the current is spread over the cross-section, so the circle encloses only I_enc=I(r²)/(a²) and B=(μ₀Ir)/(2π a²) — a straight line rising from zero on the axis. Outside the wire (r>a) the whole current is enclosed, so B=(μ₀I)/(2π r) — a hyperbola falling away. The two agree at the surface: both give B=(μ₀I)/(2π a) at r=a, which is the largest field anywhere. So the graph rises in a straight line to a peak at r=a and then falls off as 1/r, which is panel (2).

The magnetic field due to a straight long conductor of uniform cross-section of radius a and carrying a steady current is represented by ______. In each graph B is the magnitude of the field and r is the distance from the axis of the conductor.

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