Question

A 1.0-m lamp cord leads from the 110-V outlet to a lamp having a 75-W lightbulb. The cord consists of two insulated parallel wires 5.0 mm apart and held together by the insulation. One wire carries the current into the bulb, and the other carries it out.

4)Compare the field between 1.0 mm from one of the wires in the same plane in which the two wires lie with Earth’s magnetic field? (Express your answer to two significant figures.)

Answer 1

The electric power of the lamp, P = 75 W

the voltage in the circuit, V = 110 Volt

current flowing in the circuit is I.

P = V*I

I = P/V = 75/110 = 0.6818 A

Thus, the current flowing through each wire is I = 0.6818 A

Let us now assume that we want to calculate magnetic field produced at a point located at a distance, d₂ = 1.0 mm = 1*10^-3 m from the wire on the right side i.e. wire 2 such that the magnetic field produced by wire 2 at d₂ = 1.0 mm = 1*10^-3 m on right is

B₂ = u_oI/2*pi*d₂

B₂ = [4*pi*10^-7*0.6818]/[2*pi*1*10^-3] = 1.3636*10^-4 T

This field is directed outwards.

while, this point is located at a distance, d₁ = d+d₂ = (5.0+1.0)mm = 6.0*10^-3 m from the wire 1 that carries current inside the bulb such that the magnetic field produced sue to wire 1 is

B₁ = u_oI/2*pi*d₁

B₁ = [4*pi*10^-7*0.6818]/[2*pi*6.0*10^-3] = 2.27*10^-5 T directed inwards.

Thus, net magnetic field at the point of observation is B = B₁ - B₂ = 2.27*10^-5 T - 1.3636*10^-4 T

B = - 1.14*10^-4 T. i.e., the magnitude is 1.14*10^-4 T and is directed outwards.

Part 4]

The ratio of magnetic field B = 1.14*10^-4 T obtained above and earth's magnetic field B_earth = 0.5*10^-4 T is

B/B_earth =  1.14*10^-4 T /  0.5*10^-4 T

B/B_earth = 2.2727

Hence, the field between 1.0 mm from one of the wires in the same plane in which the two wires lie is 2.2727 times the Earth's magnetic field B_earth.