Compass deflection by overhead transmission lines

AI Thread Summary
A hiker's compass is affected by the magnetic field produced by a nearby overhead transmission line carrying 807A of current. The magnetic field strength directly underneath the line is calculated to be approximately 3.164*10^-5 T. To find the deflection angle of the compass, the vector sum of the Earth's magnetic field and the field from the transmission line is analyzed. By assembling the two magnetic field vectors and using trigonometric relationships, the deflection angle is determined to be around 33 degrees. This shows how overhead transmission lines can influence compass readings significantly.
SQUIDDO
Messages
4
Reaction score
0
Problem Description:
A hiker is reading a compass under an overhead transmission line that is 5.1m above the ground and carries a current of 807A in a horizontal direction from North to South. Assume the Earth's field is of the order 0.5*10^-4T.

A) Determine the magnitude of the field produced by the transmission line at a point directly underneath

B) Due to the transmission line, the compass will be deflected by a certain angle instead of pointing north. Find this angle in degreesRelevant equations:
1. Magnetic field of a straight conductor at distance 'd' = (μ/4π)*(2I/d)
Attempt at Solution:
Using the equation above and a given value for mu, I was able to find the strength of the magnetic field directly underneath the wire as being 3.164*10^-5

As for the next step, I'm totally lost. I was thinking of finding the forces on the compass due to the transmission line compared to Earth's field, and using vector lines to find the deflection, but I have no idea of the charge on the compass or anything.

I also thought I could just take the relative directions and sizes of the Earth's field and then draw vectors from there, but I don't know if that's allowed.

What can I do? Thanks for your time!
 
Physics news on Phys.org
SQUIDDO said:
I also thought I could just take the relative directions and sizes of the Earth's field and then draw vectors from there, but I don't know if that's allowed.
Yes, it's allowed :smile:. What do you get? (You don't need the "charge" of the compass. Just figure out the direction of the net magnetic field.)
 
In that case, I draw two vectors to represent the magnetic field directly under the wire due to the wire (A), and due to the Earth (B).

Assembling them head-to-tail, tan(theta) = B/A
So theta must be ~33 degrees, which is right! Thanks!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top