Question

How much current must pass through a horizontal power transmission cable in order for the magnetic field at a location 16 m directly below it to be equal to the Earth's magnetic field, which is approximately 5.0×10−5T? How much current must pass through a horizontal power transmission cable in order for the magnetic field at a location 16 m directly below it to be equal to the Earth's magnetic field, which is approximately 5.0×10−5T?

Answer 1

Given,

Magnetic field , B = 5.0× 10^-5T

Distance , d = 16 m

A current carrying wire produce a magnetic field around it. The magnitude of magnetic field produces by long current carrying wire at a distance d from the wire is given by,

B = (μ₀I)/(2Πd)

Here, μ₀ is the magnetic permebility in free space = 4Π ×10^-7

I= current through the wire.

So, I = (2Πd×B)/μ₀

or, I = (2×3.14×16× 5.0×10^-5)/(4Π×10^-7)

or, I = 4000 Ampere.

Hence the amoun of current to be passed is 4000A.