JEE MAIN - Physics (2022 - 25th June Morning Shift - No. 10)
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 :
B $$\propto$$ r2
B $$\propto$$ r
B $$\propto$$ $${1 \over {{r^2}}}$$
B $$\propto$$ $${1 \over {{r}}}$$
Explanation
$$\int {\overline B \,.\,\overline {dl} = {\mu _0}{I_{in}}} $$
$$ \Rightarrow B \times 2\pi r = {{{\mu _0}I} \over {\pi {R^2}}} \times \pi {r^2}$$
$$ \Rightarrow B \propto r$$
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