What's the Induced Voltage in a Coil Removed from a Magnetic Field?

[agilic]

Homework Statement

A coil of wire with 100 turns and a cross-sectional area of 0.1 m2 lies with its plane perpendicular to a magnetic field of magnitude 1 T . (a) Describe in your own words the meaning of magnetic flux (b) calculate the total magnetic flux passing through the coil. (c) If the coil is rapidly removed from the magnetic field in a time of 0.2s, what is the average induced voltage in the coil?

Homework Equations

Φm = B*A

The Attempt at a Solution

I know what magnetic flux is, so I'm looking for help with (b) and (c) please.

 

[Rick88]

For a coil of N turns, \Phi=NBA.
You know all the three variables, so that's an easy calculation for b).

For c), you need to use Faraday's Law of Induction \epsilon=(-)d\Phi/dt.
I put the minus in brackets because you really need the magnitude of the voltage, not its direction. But the minus should indeed be there. Again, it's just substitution :)


R.

 

[agilic]

For a coil of N turns, LaTeX Code: \\Phi =NBA.
You know all the three variables, so that's an easy calculation for b).

For c), you need to use Faraday's Law of Induction LaTeX Code: \\epsilon =(-)dLaTeX Code: \\Phi /dt.
I put the minus in brackets because you really need the magnitude of the voltage, not its direction. But the minus should indeed be there. Again, it's just substitution :)


R.

So for (b), it's just (100)(0.1^2)(1). That gives a magnetic flux of 1. Does it look like I did that right?

 

[Rick88]

So for (b), it's just (100)(0.1^2)(1). That gives a magnetic flux of 1. Does it look like I did that right?

I thought the area was 0.1 m^2. How come it's squared now?

 

[agilic]

I thought the area was 0.1 m^2. How come it's squared now?

That's weird that the question is phrased like that. What is the difference between 0.1^2 m and 0.1 m^2? And how do I set up the equation in this case?

 

[Rick88]

well, an area has the units of m^2, as you know.
So saying that the area is 0.1^2 m makes no sense.

Therefore from what you've written above, you have A=0.1 [m^2], and all you need to do is plug it into the equation in the same way as before.

So you would have 100*0.1*1 . R.

 

[agilic]

well, an area has the units of m^2, as you know.
So saying that the area is 0.1^2 m makes no sense.

Therefore from what you've written above, you have A=0.1 [m^2], and all you need to do is plug it into the equation in the same way as before.

So you would have 100*0.1*1 .


R.

Right. Sorry, I had a brain cramp. For part (C), does this look right?

(10 T*m^2 - 0) / (0.2s) = 50 V

 

[Rick88]

It indeed does.

R.