Velocity of a slider in magnetic field

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

As soon as switch S is closed , a current would flow in the circuit and the capacitor would act like a conducting wire . Force F =ilB towards right will act on the slider . There will be an acceleration but the velocity of the slider will be 0 .

Is the question correct ?

 

[cnh1995]

There will be an acceleration but the velocity of the slider will be 0 .

That would be true if self induction were not ignored.

Hint: Apply impulse-momentum theorem.

 

[Jahnavi]

That would be true if self induction were not ignored.

Hint: Apply impulse-momentum theorem.

Why would there be an impulsive force on the slider ? Since self induction is ignored there would be an immediate current flowing through the slider resulting in a force in right direction . This force will be there till current flows . It is not an impulsive force . No ?

 

[cnh1995]

Since self induction is ignored there would be an immediate current flowing through the slider resulting in a force in right direction . This force will be there till current flows . It is not an impulsive force . No ?

It is an impulsive force. The current flows for a very short time ( ideally zero, practically very small) since there is no resistance.

 

[Jahnavi]

And this large current i for a very brief duration ∆t charges the capacitor to a value Q =i∆t ?

When the question asks to ignore self inductance , it asks to ignore motional emf of magnitude Blv generated in the slider ?

 

[cnh1995]

And this large current i for a very brief duration ∆t charges the capacitor to a value Q =i∆t ?

Right.

When the question asks to ignore self inductance , it asks to ignore motional emf of magnitude Blv generated in the slider ?

No, it asks to ignore the self inductance of the loop.

 

[Jahnavi]

No, it asks to ignore the self inductance of the loop.

Aren't we neglecting motional emf developed across the ends of the slider ?

What would happen if self inductance of the loop is not ignored ?

 

[cnh1995]

Aren't we neglecting motional emf developed across the ends of the

For the initial brief time Δt, we are ignoring the motional emf ,as Δt→0.

What would happen if self inductance of the loop is not ignored ?

It will be a second-order circuit and the initial current will be zero. It gets mathematically complicated.

 

[Jahnavi]

No, it asks to ignore the self inductance of the loop.

By self induction , you mean dΦ/dt i.e rate of change of flux ?

 

[cnh1995]

By self induction , you mean dΦ/dt i.e rate of change of flux ?

Well, when the slider "starts" moving, we can't ignore the rate of change of flux i.e. the motional emf which is the result of self induction. And I think when they asked to ignore self induction, they meant ignore motional emf for that initial brief time.

What I had in mind was the self inductance (or stray inductance) of the loop when the slider is at rest, which is purely geometrical.

 

[Jahnavi]

Then you agree with later part of post#5 .No ?

I have another question quite similar to this one .

Writing KVL at any time t ,

E - iR -Bvl - Q/C =0 .

Is this the correct equation ? And more importantly is this the way to solve this problem ?

 

[cnh1995]

Then you agree with later part of post#5 .No ?

Yes. The interval Δt is so small that we can assume the velocity transition from 0 to v_initial is instantaneous.

I have another question quite similar to this one .

Ok, but did you get the answer to the first one?

 

[Jahnavi]

Option B .

 

[cnh1995]

Then you agree with later part of post#5 .No ?

I have another question quite similar to this one .

View attachment 214168Writing KVL at any time t ,

E - iR -Bvl - Q/C =0 .

Is this the correct equation ? And more importantly is this the way to solve this problem ?

You can convert it into a DE and solve for the general time domain solution, but in this case, it is not necessary.

The problem asks for the charge after a long time, which means the circuit will have reached steady state by then. What is the steady state scenario here? What can you say about the energy balance in steady state?

 

[Jahnavi]

What is the steady state scenario here? What can you say about the energy balance in steady state?

There will be no current in the circuit . Slider will move with a constant speed v towards right . A constant motional emf Bvl will develop across the slider .KVL around the loop would be E - Bvl - Q/C = 0 . Right ?

But v is not known .

Regarding the energy balance part , QE - ∫i²Rdt =(1/2)mv² + Q²/(2C) ?

But how do we calculate the second term i.e amount of electrical energy dissipated in the resistor ?

 

[cnh1995]

Edit: Erroneous post deleted.

 

[Jahnavi]

It will be better if I make a new thread for the second problem .

Thanks a lot for your help .

 

[cnh1995]

Your reasoning for the steady state is correct.

And when the two energy storing elements are different, energy associated with both of them is same.

Hmm, I take this statement back. That's not true. It is true only when there's no dissipation. Sorry for that!
Actually, you don't need energy at all.
You can solve it in a much simpler way.

First, set up the force balance equation, in terms of velocity and charge.

 

[Jahnavi]

Yeah . It is much simpler than I thought :)

Thanks again .