How large is the voltage generated?

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The discussion centers around calculating the voltage generated by a home-built generator designed by a Boulder farmer to power a television using the Earth's magnetic field. The farmer's setup includes two conductors and a sliding rail pulled by horses at 8 miles per hour, with relevant parameters such as the Earth's magnetic field strength of 53 µT and an angle of 66 degrees. Initial calculations yielded a voltage of approximately 0.001 V, which participants questioned as being too low. The correct formula for electromotive force (EMF) was highlighted, emphasizing the importance of incorporating the area into the calculations. The conversation indicates confusion over the calculations and the need for clarification on the correct approach to determine the generated voltage accurately.
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Homework Statement


A Boulder farmer, frustrated by his monthly electricity bill, decides to power his extra-
large-screen television set by using a home-built generator. He knows that the Earth
acts like a huge magnet and wants to make use of this magnetic field to generate an
electromotive force (voltage difference) for his TV. The generator is made up of two
large conductors coming out of the house, connected to a large sliding (conducting)
rail pulled by his two horses, as shown in the diagram on the left below. In Boulder
the angle of the Earth’s magnetic field, which is pointing into the ground, with the
horizontal (ground) is 66o (see diagram on the right below) and the magnitude of the
field is 53 µT (micro T). The TV acts effectively like a 4 Ω resistor in the circuit, and
the friction between the moving parts is negligible.
Length of the slide is 15 m and a width of 25 m.

The horses can pull the sliding rail with 8 miles per hour. How large is the
voltage generated across the leads to the TV?


Homework Equations



so it would be EMF = bcos(theta)*voltage*length

The Attempt at a Solution


well i got an answer of like .001 V and that just doesn't seem right but i don't really know, just wanted to check
 
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sorry got a little more with this. so i found the Magnetic force to be 3.03x10^-23 which again doesn't seem right. i did Fmag=q(v*B) so 1.6*10^-19(3.58x.000058T) So then i need to find the work done which would just be the Force x velocity but i got my work as something very low so again i just don't know if that makes sense.
 
shimizua said:

Homework Equations



so it would be EMF = bcos(theta)*voltage*length

That doesn't look right. The correct formula for EMF involves area ...
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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