Question

A solid wire of radius 10 cm carries a current of 5.0 A distributed uniformly over its cross section. Find the magnetic field B at a point at a distance (a) 2 cm (b) 10 cm and (c) 20 cm away from the axis. Sketch a graph B versus x for 0 < x < 20 cm.

Answer 1

Given:
Magnitude of current, i = 5 A
Radius of the wire, b$=10\mathrm{cm}=10\times {10}^{-2}\mathrm{m}$
For a point at a distance a from the axis,
Current enclosed, $i\text{'}=\frac{i}{\mathrm{\pi }{b}^2}\times \mathrm{\pi }{a}^2$
By Ampere's circuital law,
$\oint B.dl={\mathrm{\mu }}_{0}i\text{'}$
For the given conditions,
$$\begin{gathered} B\times 2\mathrm{\pi }a={\mathrm{\mu }}_0\frac{i}{\mathrm{\pi }{b}^2}\times \mathrm{\pi }{a}^2 \\ \Rightarrow B=\frac{{\mathrm{\mu }}_{0}ia}{2\mathrm{\pi }{b}^2}\dots \left(1\right) \end{gathered}$$
$\left(\mathrm{a}\right)a=2\mathrm{cm}=2\times {10}^{-2}\mathrm{m}$
Again, using the circuital law, we get
$$\begin{gathered} B=\frac{4\mathrm{\pi }\times {10}^{-7}\times 5\times 2\times {10}^{-2}}{2\mathrm{\pi }\times {10}^{-2}} \\ =2\times {10}^{-6}\mathrm{T}=2\mathrm{\mu T} \end{gathered}$$

(b) On putting $a=10\mathrm{cm}=10\times {10}^{-2}\mathrm{m}$ in (1), we get
B = 10 $\mathrm{\mu T}$

(c)Using the circuital law, we get
$$\begin{gathered} \oint B.dl={\mathrm{\mu }}_{0}i \\ B=\frac{{\mathrm{\mu }}_{0}i}{2\mathrm{\pi }a}=\frac{2\times {10}^{-7}\times 5}{20\times {10}^{-2}} \\ =5\times {10}^{-6}\mathrm{T}=5\mathrm{\mu T} \end{gathered}$$