Question

At what distance from a long straight wire carrying a

currentof 5.0A is the magnitude of the magnetic field due to the
wireequal to the strength of the Earth's magnetic field of about
5.0 x10^-5 T?

Answer 1

The distance is 2 cm

Solution:

According to the question:

Magnetic field of Earth, B_{E} =
`5.0* 10^(- 5) T`

Current, I = 5.0 A

We know that the formula of magnetic field is given by:

`B = \farc{\mu_(o)I}{2\pi d}`

where

d = distance from current carrying wire

Now,

`d = (\mu_(o)I)/(2\pi B)`

`d = (4\pi* 10^(- 7)* 5.0)/(2\pi* 5.0* 10^(- 5))`

d = 0.02 m 2 cm