If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.50 x 10-4 T) at a distance of 15
1 answer:
You might be interested in
Answer:
a.
b.
c.
Explanation:
First, look at the picture to understand the problem before to solve it.
a. d1 = 1.1 mm
Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:
To solve the equations we have to convert all units to those of the international system. (mm→m)
μ0 is the constant of proportionality
μ0=4πX10^-7 N*s2/c^2
b. d2=3.6 mm
Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:
J: current density
c: outer radius
b: inner radius
The cilinder's current is negative, as it goes on opposite direction than the wire's current.
c. d3=7.4 mm
Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:
As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.
