A long, straight wire of radius a carries a current distributed uniformly over its cross-section. The ratio of the magnetic
Question:
A long, straight wire of radius a carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance $\frac{a}{3}$ and $2 a$, respectively from the axis of the wire is:
Correct Option: 1
Solution:
(1) Let $a$ be the radius of the wire
Magnetic field at point $A$ (inside)
$B_{A}=\frac{\mu_{0} i r}{2 \pi a^{2}}=\frac{\mu_{0} i \frac{a}{3}}{2 \pi a^{2}}=\frac{\mu_{0} i}{\pi a^{2}} \frac{a}{6}=\frac{\mu_{0} i}{6 \pi a}$
Magnetic field at point B (outside)
$B_{B}=\frac{\mu_{0} i}{2 \pi(2 a)}$
$\therefore \quad \frac{B_{A}}{B_{B}}=\frac{\frac{\mu_{0} i}{6 \pi a}}{\frac{\mu_{0} i}{2 \pi(2 \mathrm{a})}}=\frac{4}{6}=\frac{2}{3}$