We know inside the wire $\begin{array}{ll}
B=\frac{\mu_0 I}{2 \pi R^2} \cdot r & (O < r < R) \
\text { And } B=\frac{\mu_0 I}{2 \pi r} & \text { for }(R < r)
\end{array}$
JEE Main 2025 — Physics Electromagnetism
A long straight wire of a circular cross-section with radius ' a ' carries a steady current I. The current I is uniformly distributed across this cross-section. The plot of magnitude of magnetic field B with distance r from the centre of the wire is given by
Held on 24 Jan 2025 · Verified 6 Jul 2026.




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