Question
A large water main is 2.50 m in diameter and the average water velocity is 6.00 m/s. Find the Hall voltage produced if the pipe runs perpendicular to the Earth’s 5.00×105 T5.00\times 10^{-5}\textrm{ T} field.
Question by OpenStax is licensed under CC BY 4.0
Final Answer

0.750 mV0.750\textrm{ mV}

Solution video

OpenStax College Physics, Chapter 22, Problem 22 (Problems & Exercises)

OpenStax College Physics, Chapter 22, Problem 22 (PE) video thumbnail

In order to watch this solution you need to have a subscription.

Start free trial Log in
vote with a rating of votes with an average rating of .

Calculator Screenshots

  • OpenStax College Physics, Chapter 22, Problem 22 (PE) calculator screenshot 1
Video Transcript
This is College Physics Answers with Shaun Dychko. A large water pipe with a diameter of 2.50 meters is carrying some water to the right in my picture at some speed of 6.00 meters per second and the Earth's magnetic field is perpendicular to that velocity and so I have drawn it going into the page and if you imagined your fingers going into the page along these magnetic field lines and your thumb pointing to the right in the direction of the velocity of the water, any positive charges in that water would experience a force upwards because that's the direction the palm is facing and so you are gonna have positive charges up here and then negative charges will experience a force, it's called the Lorentz Force— this magnetic field applying a force on a moving charge is the Lorentz Force— the negative charges will build up on this other side of the pipe and these charges are ions that are dissolved in the water so that's just an assumption that's being made here is that this is not pure water because absolutely pure water would have no Hall effect at all because there are no free charges to move around but if there's some salt in the water or some other minerals then there would be free charges because they dissociate into these ions. Okay! So we have a voltage then established between this charge distribution and these charges being separated like this will create an electric field and its direction will be downwards like this away from the positive charge and that would tend to oppose the force due to the Lorentz Force so you have electric field force going this way and then you have the Lorentz Force going that way on the positive charge and then likewise, you know, on these negative charges, there would be a force upwards due to the field and a force downwards due to the Lorentz Force. And these are equal and so we say that the force due to this electric field equals the Lorentz Force and the force due to the electric field is the charge multiplied by the field strength and that equals the charge times its speed times the magnetic field strength and then cancel the q's on both sides and you get the electric field is the speed times the magnetic field. We can also say that the electric field is the voltage divided by the diameter of the tube this is... when you have two surfaces that are long and parallel, you can say this about them and we can solve this for v by multiplying both sides by d and you have the voltage then is the electric field strength times the diameter of the tube but the electric field we figured out before is the speed times the magnetic field strength so we substitute that in and we have 6.00 meters per second times 5.00 times 10 to the minus 5 tesla— magnetic field strength of the Earth— times 2.50 meters—diameter of the pipe— and that works out to 0.750 millivolts. And we expected a really small number here because you never experience a shock or electrocution by touching a pipe that has moving water in it so we would expect this voltage to be very small, as it is.