Question

A copper wire of diameter 1.6 mm carries a current of 20 A. Find the maximum magnitude of the magnetic field →B due to this current.

Answer 1

A copper wire of diameter 1.6 mm carries a current of 20 A. the maximum magnitude of the magnetic field →B due to this current is B= 5 mT

Explanation:

Step 1:

Given data in the question  

I = 20 A

d = 1.6 mm = $1.6 \times 10^{-3} \mathrm{m}$

$r=0.8 \times 10^{-3} m$ ( radius is half of diameter )

I is Magnitude of current  

D is wire Diameter  

r is wire radius  

Step 2:

The intensity of the magnetic field is  

$B=\frac{\mu_{0} i}{2 \pi r}$

$B=\frac{4 \pi \times 10^{-7} \times 20}{2 \pi \times 0.8 \times 10^{-3}}$

$B=\frac{2 \times 10^{-7} \times 20}{0.8 \times 10^{-3}}$

  $=\frac{40 \times 10^{-7}}{0.8 \times 10^{-3}}$

  $=\frac{400 \times 10^{-7}}{8 \times 10^{-8}}$

  $=\frac{50 \times 10^{-7}}{10^{-3}} T$

 $\begin{aligned}&=50 \times 10^{-4} T\\&=5 \times 10^{-3} T\\&B=5 m T\end{aligned}$

The maximum magnitude of the magnetic field →B due to this current is 5 mT

Answer 2

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B =5mT

hope it help

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