Induced voltage from a rod sliding on rods in a B-Field

[cdummie]

Homework Statement

Two long rods with distance between them a=100mm, over those two rods there's another rod sliding, friction between them is negligible. The rods are placed in homogenous magnetic field whose B is equal 1T, with direction shown in the picture, between the rods there is voltage generator with constant E, and resistance R=0,1Ω. When mechanical force on rod is zero, then, velocity of rod is v₀=10m/s. Represent power of this engine in function of the velocity of the rod and find the mechanical force that acts when power of this engine is strongest, and find that power. Ignore the self-inductance (i believe self-inductance is the correct term).

Homework Equations

e[/B]_ind=∫(v×B)dl

The Attempt at a Solution

When there's no mechanical force, then there's only magnetic force due to generator E, but i don't know how to express power using velocity and how can i express it if i don't know is there mechanical force (it says at first that if there's no mechanical force then v is known), i mean is mechanical force important for finding power of the e_ind "generator" and how can i find the expression wanted?

 

[rude man]

Well, this one's a bit more involved than your farady disc.
Questions for you:
1. what is the net force on the sliding rod, since its velocity is constant?
So what are the two forces that must cancel each other as the rod slides along?
Can you determine E from this?
2. When a force is imposed counter to the direction of v, what is then the net current and force on the sliding rod? Power?
3. maximize the power the usual way.

 

[cdummie]

Well, this one's a bit more involved than your farady disc.
Questions for you:
1. what is the net force on the sliding rod, since its velocity is constant?
So what are the two forces that must cancel each other as the rod slides along?
Can you determine E from this?
2. When a force is imposed counter to the direction of v, what is then the net current and force on the sliding rod? Power?
3. maximize the power the usual way.

1. As far as i know, rod is moving only when sum of all forces that act on it are not equal zero, since mechanic force is zero, then there's only magnetic force due to generator E, and it's F_magnetic=IB×a, considering the angles between the two vectors we have F_magnetic=IBa, I=(E+e_ind)/R and i have everything but mechanical force and E here but i could find e_induced if in case when only magnetic force acts and it's e_induced=vBa but that still can't help me find E so i am not really sure what to do with this.

2. If there's a force opposite to the direction of v, then v could change the direction if the new force is stronger than the one that pushed rod before it. I am not sure for the power, how can i express power? Should it be P=I*(e_ind+E) because i don't have E yet?

3. Usual way? By taking derivative?

 

[rude man]

1. As far as i know, rod is moving only when sum of all forces that act on it are not equal zero

That's not what Newton said ...

2. If there's a force opposite to the direction of v, then v could change the direction if the new force is stronger than the one that pushed rod before it.

In that case the engine would not be providing power, would it? No, direction of v does not change after external mechanical force is applied.

I am not sure for the power, how can i express power? Should it be P=I*(e_ind+E) because i don't have E yet?

As I suggested last time, E can be expressed in terms of v (and constants).

3. Usual way? By taking derivative?

Right

 

[cdummie]

That's not what Newton said ...

But then i don't understand. What would happen if there's mechanic force equal by intensity to the magnetic force and with opposite direction?

In that case the engine would not be providing power, would it? No, direction of v does not change after external mechanical force is applied.

Again, i don't understand, why it wouldn't and why it won't change direction, if mechanical force is stronger?

As I suggested last time, E can be expressed in terms of v (and constants).

Well i know that E_ind=v×B but that's induced electric field, not E, so i am not sure about this either.

 

[rude man]

But then i don't understand. What would happen if there's mechanic force equal by intensity to the magnetic force and with opposite direction?

One thing at a time. You haven't done the first part yet, which is finding E, which you do before a mechanical external force is applied. The rod is moving to the left at 10 m/s with zero current in the sliding rod. Find E.

Again, i don't understand, why it wouldn't and why it won't change direction, if mechanical force is stronger?

You want the motor to provide work against the external force. That means the sliding rod has to move to the left, always. So the external force, applied to the right, cannot be equal to the magnetic force or then v = 0 and no work would be done. You're applying a force somewhere between zero and the mag. force, and the direction of motion is always to the left, as the v arrow shows.
If you had a motor running clockwise, would you force it to turn counterclockwise with an excessive external force? You'd be doing work on the motor, not the motor doing work on you. Same thing here.

Well i know that E_ind=v×B but that's induced electric field, not E, so i am not sure about this either.

E is the voltage of the generator, not any electric field. See above.

 

[cdummie]

One thing at a time. You haven't done the first part yet, which is finding E, which you do before a mechanical external force is applied. The rod is moving to the left at 10 m/s with zero current in the sliding rod. Find E.

Then. i could consider that there are two generators: E and e_ind so i should have I=(E+e_ind)/R , but if current is zero, then E=-e_ind. But i still don't understand how can current be zero. Does this means that every time when i have a generator and only force that acts on rod is magnetic force due to that generator, then e_ind will always be equal (by intensity) to E?

 

[rude man]

Then. i could consider that there are two generators: E and e_ind so i should have I=(E+e_ind)/R , but if current is zero, then E=-e_ind. But i still don't understand how can current be zero. Does this means that every time when i have a generator and only force that acts on rod is magnetic force due to that generator, then e_ind will always be equal (by intensity) to E?

I don't know about generalizing the way you did.
For this problem, the total force, and current, have to be zero, since the sliding rod is not accelerating.
This is of course before a mechanical force is applied towards the right. At that point, the current has to be finite to counter the mech. force. There will still be no acceleration so the net force on the rod will always be zero..

 

[cdummie]

I don't know about generalizing the way you did.
For this problem, the total force, and current, have to be zero, since the sliding rod is not accelerating.
This is of course before a mechanical force is applied towards the right. At that point, the current has to be finite to counter the mech. force. There will still be no acceleration so the net force on the rod will always be zero.

Ok, i think i understand now, basically, F_magnetic=Ia×B=0 , a is not zero, B is not zero, so current must be zero, now, I=(E+e_ind)/R =0 SO E + e_ind=0, E=-e_ind, e_ind=∫(v×B)dl =vBa=1V which means that E=-1V. Now for the second part. When mechanical force applies, magnetic force won't be zero anymore, that would mean that e_ind won't be equal by intensity to the E, how can i then express power of engine in function of velocity.

 

[rude man]

Ok, i think i understand now, basically, F_magnetic=Ia×B=0 , a is not zero, B is not zero, so current must be zero, now, I=(E+e_ind)/R =0 SO E + e_ind=0, E=-e_ind, e_ind=∫(v×B)dl =vBa=1V which means that E=-1V.

You're on the right track now! Except for the signs. Set up a coord. system: v = -v i, B = b k and a = -a j so e_ind=∫(v×B)dl = - vBa and E = + vBa. As shown on your diagram.

Now for the second part. When mechanical force applies, magnetic force won't be zero anymore, that would mean that e_ind won't be equal by intensity to the E, how can i then express power of engine in function of velocity.

OK: the sliding rod is moving to the left against a force F_mech directed to the right; how much energy does the rod have to produce to move the rod a distance x? and what is the corresponding power?

 

[cdummie]

You're on the right track now! Except for the signs. Set up a coord. system: v = -v i, B = b k and a = -a j so e_ind=∫(v×B)dl = - vBa and E = + vBa. As shown on your diagram.
OK: the sliding rod is moving to the left against a force F_mech directed to the right; how much energy does the rod have to produce to move the rod a distance x? and what is the corresponding power?

I don't know about the energy, but i know that P=F_mag*v (at least i think it should be magnetic force, since the rod has the same direction as magnetic force). Now, since F_mag=IaB, then F_mag=(E+e_ind)aBv/R. Is this correct?

 

[rude man]

I don't know about the energy, but i know that P=F_mag*v (at least i think it should be magnetic force, since the rod has the same direction as magnetic force).

Right.

Now, since F_mag=IaB, then F_mag=(E+e_ind)aBv/R. Is this correct?

Almost. Redo your last expression.
I again urge making units checking a habit. F_mag has units of force which is F whereas (E+e_ind)aBv/R has units of power which is FLT^-1. So your last equation contains an error. Maybe a typo. Really small but still all-important.

After you fixed that, how about eliminating E and e_ind with expressions in v?

 

[cdummie]

Right.
Almost. Redo your last expression.
I again urge making units checking a habit. F_mag has units of force which is F whereas (E+e_ind)aBv/R has units of power which is FLT^-1. So your last equation contains an error. Maybe a typo. Really small but still all-important.

After you fixed that, how about eliminating E and e_ind with expressions in v?

I think i know, its:

P=F_magv

now F_mag=IaB

and I=(E+e_ind)/R

so F_mag=aB(E+e_ind)/R

and P=aBv(E+e_ind)/R

or of i express e_ind with v (i believe you are asking me to do that since v, a and B are given values) then i have:

P=aBv(E+avB)/R

Now, what is the expression for mechanical force, since i have to know it's maximum (i mean it can't be the same as for magnetic, since they are not equal and only way i can express mechanic force is F_meh=IaB but i doubt it is correct since it's the same as expression for magnetic force)?

 

[rude man]

I think i know, its:

P=F_magv

now F_mag=IaB

and I=(E+e_ind)/R

so F_mag=aB(E+e_ind)/R

and P=aBv(E+e_ind)/R

or of i express e_ind with v (i believe you are asking me to do that since v, a and B are given values) then i have:

P=aBv(E+avB)/R

v isw not a "given value" now. v was given only before the mechanical force was applied. Once F_mech is applied, v will be smaller. But your formula for P is now correct if you watch your signs of E and avB.

Now, what is the expression for mechanical force, since i have to know it's maximum (i mean it can't be the same as for magnetic, since they are not equal and only way i can express mechanic force is F_meh=IaB but i doubt it is correct since it's the same as expression for magnetic force)?

Oh but they ARE equal! The rod is still not accelerating, so F_mech + F_mag = 0. So don't be a Doubting Thomas; F = iaB is correct. But watch the signs.
And you still haven't replaced E with a function of v. (Be careful, v here should really be labeled v₀ since it's = 10 m/s whereas v in your P expression is v AFTER the mech. force is applied).

 

[cdummie]

v isw not a "given value" now. v was given only before the mechanical force was applied. Once F_mech is applied, v will be smaller. But your formula for P is now correct if you watch your signs of E and avB.

I know what you mean, because it should be E-e_ind in the parentheses before i "switched" e_ind with avB (picture confused me, it shows wrong direction of the e_ind).

Oh but they ARE equal! The rod is still not accelerating, so Fmech + Fmag = 0. So don't be a Doubting Thomas; F = iaB is correct. But watch the signs.
And you still haven't replaced E with a function of v. (Be careful, v here should really be labeled v0 since it's = 10 m/s whereas v in your P expression is v AFTER the mech. force is applied).

E=avB since e_ind=-avB and E=-e_ind, now i should find derivative of F_mech=-iaB but none of these are variables, how can i do that?

 

[rude man]

E=avB since e_ind=-avB and E=-e_ind, now i should find derivative of F_mech=-iaB but none of these are variables, how can i do that?

As I said last time, you need to distinguish between v before the force, and v after. Let us forthwith define v₀ as the initial velocity, which is a constant = 10 m/s. Then, v after the force is applied IS a variable; it varies with the mechanical force magnitude.

 

[cdummie]

As I said last time, you need to distinguish between v before the force, and v after. Let us forthwith define v₀ as the initial velocity, which is a constant = 10 m/s. Then, v after the force is applied IS a variable; it varies with the mechanical force magnitude.

Well i guess i could, Since F_mech=-iaB, express B in terms of v like this E =vBa, then B=E/va so F_mech=-iaE/va=-iE/a and then find derivative of this, is this correct?

 

[rude man]

Well i guess i could, Since F_mech=-iaB, express B in terms of v like this E =vBa, then B=E/va so F_mech=-iaE/va=-iE/a and then find derivative of this, is this correct?

You're still not separating v from v₀.

 

[cdummie]

You're still not separating v from v₀.

But how can i do that?

 

[rude man]

You found the relationship between E and v₀ back in post 9! That relationship never changes, even after F_mech is applied. Whereas v is a function of F_mech.

 

[cdummie]

You found the relationship between E and v₀ back in post 9! That relationship never changes, even after F_mech is applied. Whereas v is a function of F_mech.

So E=v₀Ba, is that correct?

 

[rude man]

So E=v₀Ba, is that correct?

Correct!

 

[cdummie]

Correct!

Great, now, F_mech=-iaB and E=v₀Ba => B=E/v₀a so F_mech=-iE/v₀ but all of these are constants, how can i find derivative?

 

[rude man]

[QUOTE="cdummie, post: 5204541, member: 541888"so F_mech=-iE/v₀ but all of these are constants, how can i find derivative?[/QUOTE]
Fmech is not equal to -iE/v₀. Fmech is chosen to maximize power.
Try again; relate F_mech to F_ind , remembering that F_ind can be expressed in terms of v.
I think you need to step back and try to understand what's going on here. You have a linear motor working against a force Fmech. If Fmech is too small, power = Fmech⋅v will be small. If Fmech is too big, Fmech⋅v again will be small since v will be small (dragging the motor). So there is one and only one value for Fmech for which Fmech⋅v is maximized. But Fmech can be expressed in terms of v, so you eliminate Fmech explicitly and are left with an expression in v you need to maximize wrt v.