Question

A straight thick long wire of uniform cross-section of radius a is carrying a steady current I.Calculate the ratio of magnetic field at a point $a/2$ above the surface of the wire to that at a point $a/2$ below its surface. What is the maximum value of the field of this wire?

Answer 1

The distance of point P from the axis of wire

$=a+\frac{a}{2}=\frac{3a}{2}$

Magnetic field at point P at distance $r(=\frac{3a}{2})$ from the axis of the wire is

$BP={\mu }_{0}4\pi 2Ir={\mu }_{0}4\pi 2I\left(\frac{3a}{2}\right)=\frac{{\mu }_{0}I}{3\pi a}$

The distance of point Q from the axis of wire

$r\text{'}=a-\frac{a}{2}=\frac{a}{2}$.

Magnetic field at point Q at distance $r\text{'}(=a/2)$ from the axis of the wire is

$BQ=\frac{{\mu }_0}{2\pi }\frac{Ir}{a^2}=\frac{{\mu }_0}{2\pi }\frac{I}{a^2}\times \frac{a}{2}=\frac{{\mu }_0}{4\pi }\frac{I}{a}$

$\frac{BP}{BQ}=\frac{{\mu }_{0}I}{3\pi a}\times \frac{4\pi a}{{\mu }_{0}I}=\frac{4}{3}$

Magnetic field is maximum on the surface of wire where $r=a$. Hence,

$B_{max}=\frac{{\mu }_0}{4\pi }\frac{2I}{r}=\frac{{\mu }_{0}I}{2\pi a}$