EMF generated by a magnetic field

AI Thread Summary
An automobile with a vertical radio antenna generates an EMF of approximately 9.19 x 10^-4 V while traveling at 80 km/h in the Earth's magnetic field. The calculation uses the formula ε = Blv, with the antenna length of 0.75 m and the magnetic field's angle of 52° factored in. For determining polarity, the left-hand rule indicates that the top of the antenna becomes negatively charged while the bottom becomes positively charged. Confusion arose regarding the use of the right-hand rule, but clarification confirmed the left-hand rule applies for negatively charged particles. The discussion emphasizes the importance of correctly identifying the direction of the magnetic field and vehicle motion in EMF generation.
Parad0x88
Messages
73
Reaction score
0

Homework Statement


An automobile with a vertical radio antenna of length 75 cm travels at 80 km/h eastward. The Earth’s magnetic field has magnitude of 7x10-5 T and is directed 52o downward with respect to due North. What EMF is generated between two ends of the antenna? Comment on the polarity of EMF as well.


Homework Equations


ε = Blv
To find V in m/s: 80km/h X 1000m/km X 1h/60m X 1m/60s = 22.22 m/s
l = 0.75m
θ = 52°


The Attempt at a Solution


7x10-5 X .75 X 22.22 X sin(52°)

= 9.1925 X 10-4 V

That make sense to you guys for part A?

As for Part B, the polarity, I don't really know how to determine. I know I will have to factor the direction of the car being towards the east, and the direction of the magnetic field, but I do not understand how one will interact with the other. Any hint in the right direction would be appreciated for this part!
 
Physics news on Phys.org
Parad0x88 said:
That make sense to you guys for part A?
Yes, that looks good. In my opinion you should mention which equation you are using and use SI symbols in the equation. And as for polarity, use left-hand rule to determine which end becomes negatively charged.
 
lep11 said:
Yes, that looks good. In my opinion you should mention which equation you are using and use SI symbols in the equation. And as for polarity, use left-hand rule to determine which end becomes negatively charged.

I always use units in the assignment, it just gets confusing on the computer :D

And isn't it the right-hand rule for everything related to magnetic fields?
 
Parad0x88 said:
I always use units in the assignment, it just gets confusing on the computer :D

And isn't it the right-hand rule for everything related to magnetic fields?
No, the left-hand rule is for negatively charged particles.
 
lep11 said:
No, the left-hand rule is for negatively charged particles.

Ahh I see, and since it's going downwards, then its negatively charged, is that correct?
 
Parad0x88 said:
Ahh I see, and since it's going downwards, then its negatively charged, is that correct?

If I use the left-hand rule, I am deducing that the top of the antenna would be negative and the bottom would be positive, would that be correct?
 
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...
Back
Top