Question

A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of magnetic field at a/4 inside and a/4 outside from surface of wire is​

Answer 1

Answer:

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}∮

B

.

dl

=μ

0

I

en

here, i_{en}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}i

en

=I

πa

2

π(a/2)

2

=\frac{I}{4}

4

I

so, B . dl = \mu_0i_{en}μ

0

i

en

or, B. 2π(a/2) = \mu_0μ

0

I/4

or, B = \frac{\mu_0I}{4\pi a}

4πa

μ

0

I

....(1)

case 2 : magnetic fiend at 2a :

here, i_{en}=Ii

en

=I because total current flows through wire is I.

so, B'. dl' = \mu_0Iμ

0

I

or, B'. 2π(2a) = \mu_0Iμ

0

I

or, B' = \frac{\mu_0I}{4\pi a}

4πa

μ

0

I

... (2)

from equations (1) and (2),

\frac{B}{B'}=1

B

′

B

=1