A long straight wire with a circular cross-section having radius R , is carrying a steady current I . The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current

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Correct answer: A. B ∝ r

Using Ampere's law and noting that the steady current is uniformly distributed over the cross-section of the cylindrical wire.

$B = \left[$ $\begin{array}{ll}\dfrac{{\mu}_{0}Ir}{2\pi{R}^{2}}, r \le R \\ \dfrac{{\mu}_{0}I}{2\pir}, r > R\end{array}$ $\right]$

Thus, magnetic field $B ∝ r$ from the centre for $r < R$ .

A long straight wire with a circular cross-section having radius R...

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