Question

a long straight wire of radius a carries steady current I .The current is uniformly distributed across its cross section, The ratio of the magnetic field at a/2 and 2a from axis is

Answer 1

A long straight wire of radius ‘a’ carries steady current I. also question said “ current is uniformly distributed.”

now, magnetic field due to straight long wire at perpendicular distance from its axis is found by Ampere circuital law,

$\oint{\vec{B}.\vec{dl}}=\mu_0I_{en}$

here, $i_{en}$ is current enclosed by amperian path.

case 1 : magnetic field at a/2 :

here, $i_{en}=I\frac{\pi(a/2)^2}{\pi a^2}$=$\frac{I}{4}$

so, B . dl = $\mu_0i_{en}$

or, B. 2π(a/2) = $\mu_0$ I/4

or, B = $\frac{\mu_0I}{4\pi a}$....(1)

case 2 : magnetic fiend at 2a :

here, $i_{en}=I$ because total current flows through wire is I.

so, B'. dl' = $\mu_0I$

or, B'. 2π(2a) = $\mu_0I$

or, B' = $\frac{\mu_0I}{4\pi a}$... (2)

from equations (1) and (2),

$\frac{B}{B'}=1$

Answer 2

Answer: 1

Explanation:

Please find the explanation in the attachment. ..Hope this helps. . .