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Question

Physics Question on Amperes circuital law

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

A

Br2B∝r^2

B

BrB∝r

C

B1r2B∝\frac {1}{r^2}

D

B1rB∝\frac {1}{r}

Answer

BrB∝r

Explanation

Solution

We know that,
B.dl=μ0lin∫B.dl = μ_0l_{in}
B×2πr=μ0lπR2×πr2B \times 2\pi r = \frac {μ_0l}{\pi R^2} \times \pi r^2
Br⇒ B ∝ r

So, the correct option is (B): BrB ∝ r