What is the maximum generated voltage from the swinging metal piece?

[Unkownentity]

Homework Statement

A 0,075m long piece of metal is hanging on a spring, the piece of metal is hanging horizontally and at a right angle to the magnetic field B=0,32T. On each of the ends the piece of metal is equipped with resistant free instruments and are connected to a battery with a terminal voltage of 1,0V

a) Our ammeter measures an amount of 1,40A passing through the metal when it is hanging still. How large is the resistance?

b) The metal piece is now haning vertically and performing a repititve "swinging" motion, the readings show that a voltage is generated, the readings show between 1,38A and 1,42A. What is the maximum velocity of the piece of metal?

Homework Equations

a)

U = IR

b) ?

The Attempt at a Solution

a)

Question a was kind of straight forward where R = 1,0/1,4 which gives us R = 0,71 ohms

b)

This is where i get confused though

 

[Delphi51]

The metal moving in the magnetic field (velocity, B and length of metal all perpendicular, I would assume) will result in an induced potential V = L*v*B from one end of the metal to the other. This is due to the magnetic field causing a force on the moving electrons F = q*v*B and the force acting as an electron moves the length of the metal.

In (b) the "1,38A and 1,42A" should be "1,38V and 1,42V", right?

 

[Unkownentity]

The metal moving in the magnetic field (velocity, B and length of metal all perpendicular, I would assume) will result in an induced potential V = L*v*B from one end of the metal to the other. This is due to the magnetic field causing a force on the moving electrons F = q*v*B and the force acting as an electron moves the length of the metal.

In (b) the "1,38A and 1,42A" should be "1,38V and 1,42V", right?

Yeah they are all perpendicular to each other and your motivation clearly makes sense. Although it says 1,38A and 1,42A on my question sheet so does this produce a problem?

Thanks for taking your time and helping me out

 

[Unkownentity]

I know that somehow i need to find out the maximum amount of induced voltage and then it should be fairly straight forward, by using

e=vBl

as you stated above

 

[Unkownentity]

I guess that it is not possible to achieve an answer because there is not enough information?

 

[Delphi51]

Sorry, I had things to do. It actually is "1,38A and 1,42A"in Amperes.
So, that is just a .02 A variation from the original current. The battery must still be connected. You know the resistance, so you can figure out what additional voltage difference caused it. Work with just that additional V in your e=vBl calculation.

 

[Unkownentity]

Thanks for taking your time to help me out in really appreciate it, but according the book the maximum induced "e" is e=0,0143V and by using the formula e=vBl we get that v= 0,6m/s but it still don't understand how they got the maximum induced voltage to be 0,0143V =/

 

[Delphi51]

e = I*R = 0,02*0,71 = 0,0142 V.
Just a little off. If you keep 3 digit accuracy in your resistance calc, it does work out to 0,0143 V.

 

[Unkownentity]

Yeah I know but what confused me is how the difference in current gives us the maximum induced voltage i mean if it hadn't seen the answer i would never have figured it out =/

 

[Delphi51]

Perhaps a better way to look at it is to work directly with the 1,42 A.
What voltage causes that? V = I*R = 1,42*0,714 = 1,0143 V.
The battery provides 1 V of that, so the extra 0,0143 V must be the voltage induced by e=vBl. You could work it out from the 1,38 A just as well, but the voltage would come out 0,0143 V less than the battery voltage due to the two voltages having opposite polarity.

Using the 0,02 A difference in current is just a short cut in calculation.