A long conducting wire with a circular cross-section of radius \( a \) carries a steady current \( I \), which is evenly distributed across the entire cross-section. The ratio of magnitudes of the magnetic fields at two points at distances \( \displaystyle\frac{a}{3} \) and \( \displaystyle\frac{4a}{3} \), respectively, from the axis is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 4:9 \)
Apply Ampere’s circuital law to a circle of radius \( r \) about the axis. With the current spread uniformly over the cross-section, a circle inside the wire encloses only the fraction \( r^{2}/a^{2} \) of the current.
Inside \( (r \lt a) \):
\( \displaystyle B(2\pi r) = \mu_{0}I\frac{r^{2}}{a^{2}} \ \Rightarrow\ B = \frac{\mu_{0}Ir}{2\pi a^{2}} \)
Outside \( (r \gt a) \):
\( \displaystyle B = \frac{\mu_{0}I}{2\pi r} \)
At \( r = a/3 \) (inside):
\( \displaystyle B_{1} = \frac{\mu_{0}I(a/3)}{2\pi a^{2}} = \frac{\mu_{0}I}{6\pi a} \)
At \( r = 4a/3 \) (outside):
\( \displaystyle B_{2} = \frac{\mu_{0}I}{2\pi(4a/3)} = \frac{3\mu_{0}I}{8\pi a} \)
\( \displaystyle \frac{B_{1}}{B_{2}} = \frac{1/6}{3/8} = \frac{8}{18} = \frac{4}{9} \)
Shortcut: In units of the surface value \( B_{a} \), the field is \( B_{a}(r/a) \) inside and \( B_{a}(a/r) \) outside, so the ratio is \( \frac{1}{3} : \frac{3}{4} = 4:9 \).