Question

A transmission wire carries a current of 100 A. What would be the magnetic field B at a point on the road if the wire is 8 m above the road?

Answer 1

Explanation:

Given:Current through the transmission wire: I = 100 A

Distance between wire and the road : d = 8 mFormula used:

By Ampere’s Law for a current carrying wire is

Where,

B is the magnitude of magnetic field,

μ0 is the permeability of free space

μ0= 4π × 10-7 T mA-1

d is the distance between the current carrying wire and the required point.

Substituting we get,

B = 2.5× 10-6 T

Hence, magnitude of magnetic field at a point on the road due to current carrying transmission wire is 2.5× 10-6 T.

Answer 2

A transmission wire carries a current of 100 A. then  the magnetic field B at a point on the road if the wire is 8 m above the road is $B = 2.5 \mu T$

Explanation:

Given data in the question  :

Current magnitude  ( I ) = 100 A

Distance of the road from the wire, d = 8 m

The wire is encircled by the magnetic field created by a steady current flowing through a very long straight line. It has magnitude at a point P a radial distance from the wire .

$B= \frac{mu_0 i}{2\pi d }$

On substituting the given values, we get

=$\frac{(4\pi \times10^{-7} \times 100)}{(2\pi \times8)}$

=$\frac{(2 \times 10 ^{-7} \times 100)}{8}$

= $\frac{(200 \times 10^{-7} )}{8} T$

= $25\times 10 ^{-6} T$

$B = 2.5 \mu T$