Potential difference between two points in a loop

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

If we consider the bigger semicircular arc BD alone then potential difference between B and D will be 4BvR .

If we consider smaller semicircular arc AC alone then potential difference between A and C will be 2BvR .

If we consider straight line BA alone then potential difference between A and B will be BvR .

If we consider straight line CD alone then potential difference between C and D will be BvR .

I think option 4) looks correct but I am not too sure about the answer .

 

[BvU]

It remains a guessing game unless you apply a relevant equation ...

 

[cnh1995]

I think option 4) looks correct

I agree. You can convert the curvy loop into a single straight wire between B and D and find the emf between B and D.

 

[Jahnavi]

I agree. You can convert the curvy loop into a single straight wire between B and D and find the emf between B and D.

Thank you so much

But the problem is that answer given is option 2) .

Do @Charles Link and @gneill also agree that option 4) is the correct choice ?

 

[BvU]

What is the relevant equation you use ?

 

[Charles Link]

@Jahnavi The loop that is moving through the constant magnetic field experiences no EMF around the loop. (This part is what my calculations also showed, but I googled it, see https://www.quora.com/Is-an-EMF-induced-when-a-coil-is-moving-inside-a-constant-magnetic-field to check the result). $\\$ Meanwhile, please answer @BvU 's question. Can you write an integral equation for the EMF that you computed? It would be good to have an equation to use, in going along the wire from B to D, rather than just assuming you can make it into any shape you want to get the answer. $\\$ Also, I think there is some additional analysis that is worth considering in this problem. If there is an EMF in both paths from B to D, that implies the existence of an electric field in the wire along both paths. Why is there then no electrical current flowing along those paths?

 

[Jahnavi]

Thanks Charles for your input .

Thank you @cnh1995 .

 

[cnh1995]

It would be good to have an equation to use, in going along the wire from B to D, rather than just assuming you can make it into any shape you want to get the answer.

Say you have a straight wire between D and B. The emf induced in this wire would be 4BvR. Now if there was a circular arc between B and D, the emf would still be equal to 4BvR, because the effective length of the arc is equal to the straight-line distance between its end-points. The straight wire and the arc both cut the same area per unit time, hence the voltage between their ends should be same in both the cases.
This is why I suggested that the loop could be collapsed into a single wire between B and D.
Corrections are welcome!

 

[Charles Link]

Say you have a straight wire between D and B. The emf induced in this wire would be 4BvR. Now if there was a circular arc between B and D, the emf would still be equal to 4BvR, because the effective length of the arc is equal to the straight-line distance between its end-points. The straight wire and the arc both cut the same area per unit time, hence the voltage between their ends should be same in both the cases.
This is why I suggested that the loop could be collapsed into a single wire between B and D.
Corrections are welcome!

@cnh1995 That analysis is correct, but what I was looking for is EMF $\mathcal{E}=\int \vec{E}_{induced} \cdot d \vec{l}$, and $\vec{E}_{induced}=\vec{v} \times \vec{B}$. In doing this line integral, any x-components of $d\vec{l}$ give zero contribution, because $\vec{E}_{induced}$ points in the minus y direction, with the dot product of $\vec{E}_{induced} \cdot d \vec{l}$ giving zero when coupled to the x-component of $d \vec{l}$. $\\$ And also, why is there no current flow even with an $\vec{E}_{induced}$ in the wire? The reason is there is an immediate static buildup of charge to the lower side of the loop, which will have an electric field $\vec{E}_s$, equal and opposite $\vec{E}_{induced}$, so that $\vec{E}_{total} =\vec{E}_{induced}+\vec{E}_s=0$ in the wire. The potential, from this static charge, is then given by $V=-\int \vec{E}_s \cdot d \vec{l}=\int \vec{ E}_{induced} \cdot d \vec{l}$. If I calculated the directions properly the point of higher potential is at D. $\\$ And since the problem has already been essentially solved, it should be permissible to furnish these additional inputs. ( In general, the Homeworker Helpers are not allowed to supply the answers to the solutions, but instead the OP is required to put the effort into getting the solution).

 

[cnh1995]

@Charles Link, your explanation about zero current in the loop is displayed in a strange font (mathjax I guess).
Without using any math, it can be seen that there is no "change of flux" in the loop as it moves towards left, giving E_net=0.

 

[Charles Link]

@Charles Link, your explanation about zero current in the loop is displayed in a strange font (mathjax I guess).
Without using any math, it can be seen that there is no "change of flux" in the loop as it moves towards left, giving E_net=0.

The EMF symbol is done with mathcal{E} with a \ in front of mathcal. In any case, yes, the EMF around the loop is zero. $\\$ The explanation of the electric fields being zero is necessary if you just have a single wire in motion for explaining why there is no current in the wire. Since the EMF around the loop is zero, yes, it does tell you there will be no current flow. $\\$ In this case with the loop, there will still be a static charge build-up on the lower end of the loop to offset the $\vec{E}_{induced}$ that arises. $\\$ Edit: "Static charge build-up" may be a somewhat poor description=in more detail, the lower end of the conducting wire will be somewhat positively charged, and the upper end somewhat negatively charged to create the electric field $\vec{E}_s$ inside the wire.

 

[rude man]

I like @cnh1995 's analysis, i.e. replacing the curved parts by straight wires. Saves a bunch of integration! That way the vertical wires are seen to be two "batteries" of equal voltage but in opposite serial connection, so i=0. And the emf drop across either vertical wire must be 4BvR by the "Blv" law (which should be employed rather than faraday for moving media).

 

[Jahnavi]

@Charles Link , I apologise for not replying to your post#6 . But @rude man 's above post is exactly how I had thought in the OP to arrive at option 4) .

You said that current doesn't flow despite an EMF because there is a conservative electric field to oppose the EMF . I agree . But that reasoning would be fine if we didn't have a loop .

In this case the left and right parts act as two batteries of equal and opposing EMF's , hence no current .

 

[Charles Link]

@Charles Link , I apologise for not replying to your post#6 . But @rude man 's above post is exactly how I had thought in the OP to arrive at option 4) .

You said that current doesn't flow despite an EMF because there is a conservative electric field to oppose the EMF . I agree . But that reasoning would be fine if we didn't have a loop .

In this case the the left and right parts act as two batteries of equal and opposing EMF's , hence no current .

@Jahnavi Your solution is quite good. With these E&M problems of loops moving in magnetic fields, it can be useful to see if you can get the same answer by an alternative method. If that is the case, then it usually means that you got the correct answer. :)

 

[Jahnavi]

Thank you @Charles Link . I really appreciate that you take time to think and write such elaborate posts explaining things as best as you can . This is what brings me here at PF . It's not just about answers that one gets in the homework help , it is these enriching discussions which I like .

Just like there are guidelines for students , there are things which helpers should also understand . Not everyone is looking for easy answers .Helpers should not just write one line and then move on another thread . If someone replies he/she should be willing to take the other person across the shore

Too much for the day .