- #1

Kavya Chopra

- 31

- 2

- Homework Statement
- A metal rod of length L, connected to a vertical shaft is rotating in a horizontal plane with a constant angular velocity $omega$ (anticlockwise as seen from above) about one of its fixed ends. Find the potential difference between the ends of the rod. Now, a uniform magnetic field in vertically upward direction is switched on. What should be the magnitude of said field so that the potential difference between the ends of the rod doubles?

- Relevant Equations
- Centripetal force=mw^2x

The first part of the problem seems easy enough, the free electrons in the wire would move in a circle owing to an electric field that would be induced in the rod which would provide the centripetal force for the same (Please correct me if I am wrong). So we have $$eE=mω^2x$$, where e is the electronic charge, m is the mass of the electron, and E is the electric field at a distance of x from the axle. So, to find the potential difference, I integrate the E over the length of the rod, to get the potential difference as $$mω^2L^2/2e$$, where the free end is at a lower potential difference.

Now for the second part, I figure that the centripetal force is now provided by the resultant Lorentz force, but since the magnetic force is also towards the centre we should have

$$eBωx−eE=mω^2x$$

Now clearly, since the potential difference doubles, so does the electric field, but I believe that in this case, the direction of electric field is the opposite, so I integrate both sides with respect to dx, and I get $$B=3mω/e$$ but the answer given is $mω/e$. I'd like to know what is incorrect about my approach, and if there are any other effects that I am neglecting in my solution, as the intended solution simply equates the motional emf to the difference of the required potential difference in the second and first part. I'd also like a bit more insight into if there are any tangential electric fields in the rod as well, or any other fields that my solution is missing.

Now for the second part, I figure that the centripetal force is now provided by the resultant Lorentz force, but since the magnetic force is also towards the centre we should have

$$eBωx−eE=mω^2x$$

Now clearly, since the potential difference doubles, so does the electric field, but I believe that in this case, the direction of electric field is the opposite, so I integrate both sides with respect to dx, and I get $$B=3mω/e$$ but the answer given is $mω/e$. I'd like to know what is incorrect about my approach, and if there are any other effects that I am neglecting in my solution, as the intended solution simply equates the motional emf to the difference of the required potential difference in the second and first part. I'd also like a bit more insight into if there are any tangential electric fields in the rod as well, or any other fields that my solution is missing.