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

Explanation:

To find: The magnetic field B at a point at a distance

Given data in the question  

Current Magnitude, i = 5 A

Wire Radius $\mathrm{b}=10 \mathrm{cm}=10 \times 10^{-2} \mathrm{m}$

For a point a away from the pole,

Enclosed Current,

$i^{\prime}=\frac{i}{\pi b^{2}} \times \pi a^{2}$

By rule of the circuit of Ampere,

$\oint B . d l=\mu_{0} t^{\prime}$

For the conditions given,

$B \times 2 \pi a=\mu_{0} i \frac{t}{\pi b^{2}} \times \pi a^{2}$

$B=\frac{\mu_{0} i a}{2 \pi b^{2}} \ldots e q^{n}(1)$

(a)  Given  $a=2 \mathrm{cm}=2 \times 10^{-2} \mathrm{m}$

We use the Circuit Act to get

$B=\frac{4 \pi \times 10^{-7} \times 5 \times 2 \times 10^{-2}}{2 \pi \times 10^{-2}}=2 \times 10^{-6} T=2 \mu T$

(b) Given $a=10 \mathrm{cm}=10 \times 10^{-2}$

On putting the value of  $a=10 \mathrm{cm}=10 \times 10^{-2}$ m in equation (1), we get  

$\mathrm{B}=10 \mu T$  

(c) We use the Circuit Act to get

$\int B . d l=\mu_{0} i$

$B=\frac{\mu_{0} t}{2 \pi a}=\frac{\mu_{0} t}{2 \pi a}$

   $=\frac{2 \times 10^{-7} \times 5}{20 \times 10^{-2}}$  

   $=5 \times 10^{-6} T$

   $=5 \mu T$