Bent rod rotating in a magnetic field

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

I am considering the two rods separately .Considering QP alone EMF induced will be (1/2)Bωl² with Q at higher potential .

Considering QR alone EMF induced will be (1/2)Bωl² with Q at higher potential .

This gives zero potential difference between P and R.

But this is incorrect .

What is the mistake ?

 

[haruspex]

Considering QR alone EMF induced will be (1/2)Bωl²

Why?
Consider six points, B, R, S..., arranged around P as corners of a hexagon. What is the potential difference between each and the next?

 

[Jahnavi]

I see my mistake .

Thanks !

 

[Jahnavi]

@haruspex , @Charles Link If I connect P and R with a rod/wire such that PQR is a complete loop , there would be no current in the loop . Right ?

So basically V_P - V_R = -(1/2)Bωl² irrespective of whether there is an incomplete loop (like in the OP) or a complete triangular loop (like the modified setup in the above para ). Right ?

 

[Charles Link]

@Jahnavi I need a few minutes to work outt what the sign of the voltage is, etc., but this problem reminded me of a previous one that I looked up, because I remembered it as having the necessary concepts to determine the answer: https://www.physicsforums.com/threa...een-two-points-in-a-loop.943126/#post-5966332 See in particular post 9 of that thread. $\\$ And yes, I think you have this one correct. $\\$ And QR above has zero voltage, because you could draw an arc of a circle connection between Q and R that has is always at a distance $r$ from P. The electric field $\vec{ E}_{induced }$ would be perpendicular to the arc at any point, because $\vec{v}$ is along the arc. (Using $\vec{E}_{induced}=\vec{v} \times \vec{B}$). This makes $\int \vec{E}_{induced} \cdot d \vec{l}=0$ because the dot product of $\vec{E}_{induced} \cdot d \vec{l}=0$ along that arc. (You arrived at that same conclusion by considering the voltage for the straight line path from points P to R).

 

[Jahnavi]

OK .

Suppose we join P and R so that we have a triangular loop. There will be equal EMF's induced in PQ and PR . There will be no induced EMF in QR .

Is this situation equivalent to as if we had a battery of EMF E between P and Q with Q at positive potential of the battery . A battery of same EMF E between P and R with R at positive potential of battery . Q and R are connected by a plain connecting wire .

So , basically the loop of three rods (modified problem ) is equivalent to two batteries of equal EMF ( between PR and PQ) placed parallel to each other . Because of this no current flows between Q and R . Or , no current flows in the loop .

Is this correct ?

 

[Charles Link]

Yes, a battery is a source of EMF, and yes, you have it correct. :) $\\$ And when two equal voltage batteries are connected in parallel, no current flows between them.

 

[Jahnavi]

OK . I am bit confused with induced EMF's and potential differences .

Suppose we change the geometry of the triangular loop such that angle between PQ and QR is 90° instead of 60° as given in the OP . PQ = QR = l . PR =√2l

Now EMF induced between P and R will be Bωl² with R at higher potential . EMF induced between P and Q will be (1/2)Bωl² with Q at higher potential .

Now why is it that despite R being at higher potential than Q , no current flows between R and Q ?

Is it because , now there will also be an induced EMF between Q and R with R at higher potential ?

Now the situation is as if there is a battery of EMF E ( between P and Q) in series with battery of EMF E (between Q and R ) .The equivalent EMF being 2E .

Just like the previous case , now this battery of EMF 2E between P and R ( via Q) will be in parallel with a battery of EMF 2E ( directly between P and R ) .

Is this thinking okay ?

 

[Charles Link]

OK . I am bit confused with induced EMF's .

Suppose we change the geometry of the triangular loop such that angle between PQ and QR is 90° instead of 60° as given in the OP . PQ = QR = l .

Now EMF induced between P and R will be Bωl² with R at higher potential . EMF induced between P and Q will be (1/2)Bωl² with Q at higher potential .

Now why is it that despite R being at higher potential than Q , no current flows between R and Q ?

Is it because , now there will also be an induced EMF between Q and R with R at higher potential ?

Now the situation is as if there is a battery of EMF E ( between P and Q) in series with battery of EMF E (between Q and R ) .The equivalent EMF being 2E .

Just like the previous case , now this battery of EMF 2E between P and R ( via Q) will be in parallel with a battery of EMF 2E ( directly between P and R ) .

Is this thinking okay ?

That is correct. The distance PR will be $s=\sqrt{2} r$ making $s^2$ a factor of 2 in the formula for the voltage between P and R. The voltage QR is most easily computed in the way that you just did . An evaluation of $\mathcal{E}_{QR}=\int\limits_{Q}^{R} E_{induced} \cdot dl$ would take a little work.

 

[Jahnavi]

OK .

So basically while dealing with induced EMF's in a complete/incomplete loop we could think of the situation in terms of batteries in various parts of the loops .

If we traverse the loop starting a point and ending at the same point , all the batteries will be in series . If there is no net induced EMF in the loop , then net equivalent EMF of all the batteries (in series ) will be zero .

Right ?

 

[Jahnavi]

Thank you very much !