Magnetic Field Produced by Power Line

[GarthVader]

1. You are standing directly under a high-voltage power line that is 4.5m above you, and a detailed map that you have indicates that the power line runs exactly north to south. Your compass needle makes an angle of 32 degrees to the east with respect to the power line's direction.

a)Knowing that the magnitude of the Earth's magnetic field is approximately 5.0*10^-5 T, determine the magnitude and direction of the magnetic field produced by the power line at ground level.
b)How much current is being carried by the power line?
c)Is the current flowing toward the north or toward the south?



2. B = μoI/2πr



3. The basic set up is where I'm stuck. I can answer questions B and C rather easily once I know A, but I'm am at a standstill on how to acquire the B field for the power line. Any help would be greatly appreciated!

 

[Sethric]

Well your compass aligns with the magnetic field, which is a vector. You know the Earth's magnetic field, and you should know which direction that points. The only other magnetic component is the field from the power line. We know the direction the current is moving, so you should know the direction of the magnetic field lines from the power line. Therefore, you know the final magnetic field vector, one of the components of that vector, and the direction of the other component. Trigonometry should give you the magnitude. And you are correct in using Ampere's Law for that.

 

[GarthVader]

So by using the law of signs, I would get:

(5x10^-5)/sin(32°) = B/sin(58°)

Solving for B I get 8x10^-5

Is this right? The magnitude of B seems high to me. Shouldn't it be smaller than the Earth's field?

 

[GarthVader]

Hang on, I had that mixed up. It should be more like:

(5x10^-5)/sin(58°) = B/sin(32°)

Which would make B = 3.1x10^-5 T

Much more reasonable.

 

[rwisz]

a) To determine the magnitude and direction of the magnetic field produced by the power line at ground level, we can use the equation B = μoI/2πr, where B is the magnetic field, μo is the permeability of free space (4π * 10^-7 T*m/A), I is the current, and r is the distance from the power line. In this case, r = 4.5m. We also know that the compass needle makes an angle of 32 degrees to the east with respect to the power line's direction, which means that the magnetic field is at a 32 degree angle to the compass needle. Using trigonometry, we can find that the magnitude of the magnetic field is B = 5.0*10^-5 T * tan(32) = 2.6*10^-5 T. The direction of the magnetic field is perpendicular to the power line, so it is pointing towards the east.

b) To determine the current being carried by the power line, we can rearrange the equation B = μoI/2πr to solve for I. Plugging in the values we know, we get I = 2πrB/μo = 2π * 4.5 * 2.6*10^-5 / 4π * 10^-7 = 2.3*10^2 A. This means that the power line is carrying a current of 230 A.

c) The direction of the current can be determined by using the right hand rule. If you point your thumb in the direction of the current (north to south), your fingers will curl in the direction of the magnetic field (east). Therefore, the current is flowing towards the south.