Electric field in a rotating rod in a magnetic field

[Kavya Chopra]

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.

 

[osilmag]

Why would the direction of the electric field be opposite of the magnetic field? Lorentz law is $F= qE + qvB$.

 

[Kavya Chopra]

@above as otherwise the lorentz force would exceed the centripetal force

 

[TSny]

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

I agree with your analysis and the result $B=3mω/e$.

I found it helpful to use vector notation in order to keep track of directions and signs. Here is a long-winded version of your analysis:

$$\sum \vec F = m \vec a$$
$$q\vec E + q \vec v \times \vec B = m(-\omega^2 r \, \hat r)$$
Here, $r$ is your $x$ and $\hat r$ is a unit vector pointing radially outward from the axis of rotation.

Since $\vec v \times \vec B = vB \hat r = \omega r B \hat r$, we have $$q\vec E + q \omega B r \, \hat r =- m \omega^2 r \, \hat r$$
We can write the charge of the electron as $q = -e$ where $e$ is the elementary charge ($e$ is a positive number).

Solving for $\vec E$, $$\vec E = \left( \frac{m \omega^2 r }{e} - B \omega r \right) \hat r$$

So, for $B = 0$, $$\vec E = \frac{m \omega^2 r}{e} \hat r$$ For $B = 0$ the electric field inside the rod points radially outward.

Inspection of $\vec E = \left( \frac{m \omega^2 r }{e} - B \omega r \right) \hat r$ shows that as $B$ is increased from zero, $\vec E$ decreases in magnitude and becomes zero when $B = m \omega /e$. Increasing $B$ beyond this value causes $\vec E$ to increase in the radially inward direction.

When $B = 3 m \omega/e$, the magnitude of $\vec E$ for a particular value of $r$ is twice the magnitude of $\vec E$ for the case where $B = 0$.

So $|\Delta V|$ for $B = 3 m \omega / e$ will be twice that for $B = 0$, but the polarity of $\Delta V$ will be switched.