Question

#12AWG copper wire (0.081” diameter) has a current carrying capacity of 20A. (a)With 20 A in the wire, what is the magnitude of the magnetic field the surface of the wire? (b) If a very long, straight piece of this wire is placed in the Earth’s magnetic field at sea level (~ 0.5 Gauss)with the wire at right angles to the field, where are the points at which the Earth’s magnetic field is canceled, i.e. the net magnetic field is zero?

Answer 1

Given Data

Diameter D = 0.081 inches = 2.058 mm so approx 2.06 mm

Therefore radius R = 1.03 mm

Subpart a) A wire is carrying 20A current so, I = 20 A

We have asked to calculate the magnitude of the magnetic field on the surface of the wire. So let's consider a circular loop of radius 1.03 mm around the wire. Therefore

Hence, With 20 A in the wire, the magnitude of the magnetic field on the surface of the wire is 3.883 x 10-3  T

Subpart b) Now we have to find the points at which the Earth’s magnetic field is cancelled, i.e. the net magnetic field is zero.

Given data

current-carrying capacity I = 20 A

Let the earth magnetic field tobe B and R be the distance of the point from the centre of the wire where the magnitude of the magnetic field is equal to the earth field.

B = 0.5 Gauss = 5 x 10-5 T

Therefore,

hence, If a very long, straight piece of this wire is placed in the Earth’s magnetic field at sea level (~ 0.5 Gauss)with the wire at right angles to the field,   the points at a distance of 8 cm from the centre of the current-carrying wire will cancel the earth's magnetic field i.e. the net magnetic field is zero