Magnetic Effects & Magnetism

JEE Main level, single correct, medium

62 of 276

Question 62 of 276

A long straight wire with a circular cross-section having radius RR, is carrying a steady current II. The current II is uniformly distributed across this cross-section. Then the variation of magnetic field due to current II with distance r(r<R)r\left(r\lt R\right) from its centre will be

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Answer: (A) B∝rB\propto r

Using Ampere's law and noting that the steady current is uniformly distributed over the cross-section of the cylindrical wire.B=[μ0Ir2πR2,r≤Rμ0I2πr,r>R]B=\left[\begin{matrix}\dfrac{{\mu }_{0}Ir}{2\pi {R}^{2}},r\leq R \\ \dfrac{{\mu }_{0}I}{2\pi r},r\gt R\end{matrix}\right]
Thus, magnetic field B∝rB\propto r from the centre for r<Rr\lt R.

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