Question

Two long parallel wires are separated by forty centimeters and carry oppositely-directed currents of ten amperes. Find the magnitude of the magnetic field (expressed in microtesla) in the plane of the wires at a point that is twenty centimeters from one wire and sixty centimeters from the other.

Answer 1

1.04μT

Step-by-step explanation:

Due to both wires have opposite currents, the magnitude of the total magnetic field is given by

`B_T=(\mu_o I)/(2 \pi r_1)-(\mu_o I)/(2 \pi r_2)`

I: electric current = 10A

mu_o: magnetic permeability of vacuum = 4pi*10^{-7} N/A^2

r1: distance from wire 1 to the point in which B is measured.

r2: distance from wire 2.

The distance between wires is 40cm = 0.4m. Hence, r1=0.2m r2=0.6m

By replacing in the formula you obtain:

`B_T=((4\pi *10^(-7)N/A^2)(10A))/(2\pi)((1)/(0.4m)-(1)/(0.6m))=1.04*10^(-6)T =1.04\mu T`

hence, the magnitude of the magnetic field is 1.04μT