Using Ampere's law and noting that the steady current is uniformly distributed over the cross-section of the cylindrical wire.B=[2πR2μ0Ir,r≤R2πrμ0I,r>R]
Thus, magnetic field B∝r from the centre for r<R.
JEE Main 2022 — Physics Electromagnetism
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 I with distance r(r<R) from its centre will be
Held on 25 Jun 2022 · Verified 6 Jul 2026.
B∝r
B∝r1
B∝r2
B∝r21
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