Calculating the induced voltage

[Biker]

Homework Statement

Excuse me for the bad drawing:
A conductor moving with a speed of 2.5 m/s in an uniform magnetic field (1.2 T) as declared in the picture below. Calculate the voltage induced in the rod.

Homework Equations

E = vBL
W = F d cos[/B]

The Attempt at a Solution

The resulting magnetic force in the -y direction and now I just need to find the voltage
## E = \frac{qvBL sin(\theta)}{q} ##
which is
## E = vBL sin(\theta) ##
## E = 2.54567 \text{volts} ##

The same book gave me an answer without considering sin(theta). The problem that book has some pretty good ideas sometimes and for the easy questions I usually find mistakes.

It even asked me to substitute E in faradays law to get the change in magnetic flux which makes no sense because it is not a closed loop

 

[CWatters]

If B is out of the page (z axis) why does the angle matter?

 

[Biker]

If B is out of the page (z axis) why does the angle matter?

But when you find the force which is
## F = qvB## it is pointing downward in the -y direction
So I can't just find the work by multiplying the force times the distance.
instead I use this
## W = F d cos(90-37)## which is equivalent to
## W = Fd sin(37) ##
Then I can find the voltage

 

[CWatters]

Ok have a think about this..

The voltage depends three things (at least these three)...

1)The direction of the B field
2)The angle of the rod in relation to the B field
and
3)The direction the rod is moving in relation to the B field

If I keep those things the same but change the coordinate system the voltage shouldn't change right?

So in the drawing below I have changed/moved the axis only. I've not changed any of those three things that effect the voltage. However the angle wrt the x-axis has changed, it's not 37 degrees anymore but the voltage should be the same right?.

 

[CWatters]

Check this page and note that the angle which matters is the angle between the velocity vector and the magnetic field. Not the angle of the velocity vector in any particular coordinate system or to any particular axis.

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/genwir2.html

 

[Biker]

I know that the magnetic force depend on the angle between the velocity vector and the magnetic field. I didn't use the angle in F = qvB sin theta
because theta is always 90 in this situation but I have used it in finding the voltage
The red arrow represents the force, The orange arrow represents the vector of the length and the angle between them is 90-37 degrees
Now W = F d cos theta
W = F L cos (90-37)
W = F L sin(37)
W = qvBL sin(37)

E = (qvBL sin(37))/q
E = vBL sin37

I don't care about the component of the magnetic force that creates a voltage across the thickness of the wire, All I care about is the component of the force that is parallel to the length of the rod

 

[CWatters]

Yes ok I think you are correct.

 

[Biker]

Yes ok I think you are correct.

Thank you CWatters :D

 

[ehild]

It even asked me to substitute E in faradays law to get the change in magnetic flux which makes no sense because it is not a closed loop[/QUOTE]

Faraday's Law states that the induced emf in a closed loop is ## E=-dΦ/dt ## where Φ is the flux across the area enclosed by the loop.
The loop does not need to be "real", made of wire. It means that bringing a unit positive charge round the loop, you need work equal to E in magnitude.
If you feel uncomfortable with a mathematical loop, you can imagine that you add wires and a voltmeter, to make a physical loop.
In your problem, imagine that the rod slides on two rails, and there is a voltmeter connected across the rails some distance away. The rod and the rails and the voltmeter make a closed loop. The area of the loop changes as the rod moves on the rails.

The magnetic field is perpendicular to the page. How much does the flux change in one second? It is the area of the shaded region multiplied by B.

 

[Biker]

It even asked me to substitute E in faradays law to get the change in magnetic flux which makes no sense because it is not a closed loop

Faraday's Law states that the induced emf in a closed loop is ## E=-dΦ/dt ## where Φ is the flux across the area enclosed by the loop.
The loop does not need to be "real", made of wire. It means that bringing a unit positive charge round the loop, you need work equal to E in magnitude.
If you feel uncomfortable with a mathematical loop, you can imagine that you add wires and a voltmeter, to make a physical loop.
In your problem, imagine that the rod slides on two rails, and there is a voltmeter connected across the rails some distance away. The rod and the rails and the voltmeter make a closed loop. The area of the loop changes as the rod moves on the rails.
View attachment 107435

The magnetic field is perpendicular to the page. How much does the flux change in one second? It is the area of the shaded region multiplied by B.

Yea, I am familiar with the moving rod example.

Hmm, So you say that for every situation you can make a mathematical loop that behaves the same as if you just use lorentz force right?
The question is then asking if you were to form a loop what would the change in flux be right?

I wonder what a loop with A = 0 will look like :/

 

[ehild]

Yea, I am familiar with the moving rod example.

Hmm, So you say that for every situation you can make a mathematical loop that behaves the same as if you just use lorentz force right?
The question is then asking if you were to form a loop what would the change in flux be right?

I wonder what a loop with A = 0 will look like :/

The loop changes because of the moving rod. The other parts of the loop are stationary. The change of area is the area swept by the moving rod.
If you connect the ends of the rod with a wire of equal length of the rod, they enclose zero area.