Question

A long, East-West-oriented power cable carrying an

Answer 1

Answer:

200A

Explanation:

Given that

the distance between earth surface and power cable d = 8m

when the current is flowing through cable , the magnitude flux density at the surface is 15μT

when the current flow throught is zero the magnitude flux density at the surface is 20μT

The change in flux density due to the current flowing in the power cable is

B = 20μT - 15μT

B =5μT -----(1)

The expression of magnitude flux density produced by the current carrying cable is

$B=\frac{\mu_0I}{2\pi d}$-----(2)

Substitute the value of flux density

B from eqn 1 and eqn 2

$\frac{\mu_0I}{2\pi d}=5\times 10^-^6\\\\\frac{(4\pi \times 10^-^7)I}{2 \pi (8)} =5\times 10^-^6\\\\I=200A$

Therefore, the magnitude of current I is 200A