The magnetic field produced by a very long straight conducting wire of radius 'a' and carrying current I is B, then the graph of magnetic field (B) → distance (r) (perpendicular to the axis of the wire) is .

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(2) (2). Given: a solid wire of radius a carrying a steady current I spread over its cross-section; B is wanted against the distance r from the axis. Idea: apply Ampere's circuital law to a circle of radius r about the axis, ∮⃗B· d⃗l=μ₀ I_enc, which gives B(2π r)=μ₀ I_enc. Inside, r<a: the circle encloses only the fraction (r²)/(a²) of the current, so B=(μ₀ I r)/(2π a²) — a straight line rising from zero at the axis. Outside, r>a: the whole current is enclosed, so B=(μ₀ I)/(2π r) — a curve falling as 1/r. The two agree at r=a, where B is greatest, (μ₀ I)/(2π a). The graph that climbs as a straight line to a peak at r=a and then decays as 1/r is panel (2).

The magnetic field produced by a very long straight conducting wire of radius 'a' and carrying current I is B, then the graph of magnetic field (B) → distance (r) (perpendicular to the axis of the wire) is ________.

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