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A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B' at radial distances $\frac{a}{2}$ and 2a respectively, from the axis of the wire is
A
$\frac{1}{2}$
B
1
C
4
D
$\frac{1}{4}$
Explanation
Inside B ∝ r, outside B ∝ 1/r; at a/2 and 2a they are equal.
Detailed Solution
Ampere's circuital law: $\oint\vec{B}\cdot d\vec{l} = \mu_0I_{enclosed}$
Inside (r = a/2): $I' = \frac{I}{\pi a^2}\times\frac{\pi a^2}{4} = \frac{I}{4}$; $B\cdot2\pi\frac{a}{2} = \mu_0\frac{I}{4} \Rightarrow B = \frac{\mu_0I}{4\pi a}$ ...(i)
Outside (r = 2a): $B'\times2\pi(2a) = \mu_0I \Rightarrow B' = \frac{\mu_0I}{4\pi a}$ ...(ii)
$\frac{B}{B'} = 1$
Inside (r = a/2): $I' = \frac{I}{\pi a^2}\times\frac{\pi a^2}{4} = \frac{I}{4}$; $B\cdot2\pi\frac{a}{2} = \mu_0\frac{I}{4} \Rightarrow B = \frac{\mu_0I}{4\pi a}$ ...(i)
Outside (r = 2a): $B'\times2\pi(2a) = \mu_0I \Rightarrow B' = \frac{\mu_0I}{4\pi a}$ ...(ii)
$\frac{B}{B'} = 1$