A steady current I flows through a long straight wire of radius 'a'. The current is uniformly distributed across its cross section. The ratio of the magnetic fields due to the wire at distance \( \displaystyle\frac{a}{4} \) and 3a respectively from the axis of the wire is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3 : 4
Ampere's circuital law \( \oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\mathrm{enc}} \) applied to a circle of radius \( r \) about the axis gives the field. Inside the wire only the current enclosed by the circle counts; outside, all of \( I \) is enclosed.
Inside (\( r \lt a \)): the enclosed current is \( I r^{2}/a^{2} \).
\( \displaystyle B\,(2\pi r) = \mu_0 I\,\frac{r^{2}}{a^{2}} \;\Rightarrow\; B_{\mathrm{in}} = \frac{\mu_0 I r}{2\pi a^{2}} \)
\( \displaystyle B\left(\frac{a}{4}\right) = \frac{\mu_0 I\,(a/4)}{2\pi a^{2}} = \frac{\mu_0 I}{8\pi a} \)
Outside (\( r \gt a \)):
\( \displaystyle B_{\mathrm{out}} = \frac{\mu_0 I}{2\pi r} \;\Rightarrow\; B(3a) = \frac{\mu_0 I}{6\pi a} \)
\( \displaystyle \frac{B(a/4)}{B(3a)} = \frac{1/8}{1/6} = \frac{6}{8} = \frac{3}{4} \)
So the ratio is 3 : 4.