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Snezhnost [94]
3 years ago
5

Two long parallel wires are separated by forty centimeters and carry oppositely-directed currents of ten amperes. Find the magni

tude 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.
Physics
1 answer:
Softa [21]3 years ago
4 0

Answer:

1.04μT

Explanation:

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

B_T=\frac{\mu_o I}{2 \pi r_1}-\frac{\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=\frac{(4\pi *10^{-7}N/A^2)(10A)}{2\pi}(\frac{1}{0.4m}-\frac{1}{0.6m})=1.04*10^{-6}T =1.04\mu T

hence, the magnitude of the magnetic field is 1.04μT

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