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

Magnetics PYQ — JEE Main 2007 · Shift 0 · Q74 with Solution

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Correct answer: D

For a long straight wire of radius $a$ carrying total current $i$ uniformly distributed over its cross-section, use Ampere’s law.

For a point at distance $r<a$ from the center,

B(2\pi r)=\mu_0 I_{\text{enc}}

where the enclosed current is proportional to enclosed area:

I_{\text{enc}}=i\cdot \frac{\pi r^2}{\pi a^2}=i\frac{r^2}{a^2}

So,

B=\frac{\mu_0 I_{\text{enc}}}{2\pi r} =\frac{\mu_0 i}{2\pi r}\cdot \frac{r^2}{a^2} =\frac{\mu_0 i r}{2\pi a^2}

Thus, at $r=\frac a2$ ,

B\left(\frac a2\right)=\frac{\mu_0 i}{2\pi a^2}\cdot \frac a2 =\frac{\mu_0 i}{4\pi a}

For a point at distance $r>a$ ,

B=\frac{\mu_0 i}{2\pi r}

Thus, at $r=2a$ ,

B(2a)=\frac{\mu_0 i}{2\pi(2a)}=\frac{\mu_0 i}{4\pi a}

\frac{B(a/2)}{B(2a)} =\frac{\mu_0 i/(4\pi a)}{\mu_0 i/(4\pi a)}=1

The correct option is:

\boxed{D: 1}

Stored correct answer: $D$

Our derived answer is also $D$ , so they agree.