A coaxial cable consists of an inner wire of radius 'a' surrounded by an outer shell of inner and outer radii 'b' and 'c' respectively. The inner wire carries an electric current i0, which is distributed uniformly across cross-sectional area. The outer shell carries an equal current in opposite direction and distributed uniformly. What will be the ratio of the magnetic field at a distance x from the axis when (i) x < a and (ii) a < x < b ?

Magnetics PYQ — JEE Main 2021 · 27 Aug · Shift 2 · Q54 with Solution

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Correct answer: A

Using Ampere’s law,

B(2\pi x)=\mu_0 I_{\text{encl}}

B=\frac{\mu_0 I_{\text{encl}}}{2\pi x}.

We need the ratio of magnetic fields at distance $x$ from the axis in two cases:

Let these fields be $B_1$ and $B_2$ respectively.

Inside the inner conductor, current is uniformly distributed.

So the enclosed current is proportional to enclosed area:

I_{\text{encl}}=i_0\frac{\pi x^2}{\pi a^2}=i_0\frac{x^2}{a^2}.

Hence,

B_1=\frac{\mu_0}{2\pi x}\left(i_0\frac{x^2}{a^2}\right) =\frac{\mu_0 i_0 x}{2\pi a^2}.

Here the Amperian loop encloses the entire inner conductor current $i_0$ , but none of the outer shell current yet.

Thus,

I_{\text{encl}}=i_0.

So,

B_2=\frac{\mu_0 i_0}{2\pi x}.

The ratio of magnetic field in case (i) to that in case (ii) is

\frac{B_1}{B_2} =\frac{\frac{\mu_0 i_0 x}{2\pi a^2}}{\frac{\mu_0 i_0}{2\pi x}} =\frac{x^2}{a^2}.

\frac{B(x<a)}{B(a<x<b)}=\frac{x^2}{a^2}

which corresponds to Option A.

Stored correct answer: A

Our derived answer: A

So they agree.