Question

Two parallel long straight wires are oriented in a north-south direction at the same height, 50.0 cm apart, and carrying currents of 8.00 AM in the same direction. Find the strength of the magnetic field (a) midway between the wires, (b) at a point in the plane of the wires but 3.00 m from the nearer of the two, and (c) at a point between and above the two wires, 1.01 m from each one.

Answer 1

a. OT

b. 9.9*10^{-7}T

c. 4.6*10^{-7}T

Step-by-step explanation:

The strength of a magnetic field generated by a current in a wire is

`B=(\mu_0I)/(2\pi r)`

where mu0 is the magnetic permeability, I is the current and r is the distance in which we want to know the magnitude of B.

a. midway between the wires

r1=0.25m

r2=0.25cm

`B=B_1-B_2=(\mu_0I)/(2\pi)((1)/(r_1)-(1)/(r_2))=0T`

b. at a point in the plane of the wires but 3.00 m from the nearer of the two

r1=3.5m

r2=3m

in this case we have

`B=B_1+B_2=(\mu_0I)/(2\pi)((1)/(r_1)+(1)/(r_2))=9.9*10^(-7)T`

c. at a point between and above the two wires, 1.01 m from each one

- between the wires, as in the point a B=0T

- above each wire:

r1=1.01m

`r_2=√((1.01m)^2+(0.5m)^2)=1.43m`

`B=B_1+B_2=4.6*10^(-7)T`

HOPE THIS HELPS!!