Induced emf in a current loop


by mattst88
Tags: current, induced, loop
mattst88
mattst88 is offline
#1
Dec2-07, 06:36 PM
P: 29
1. The problem statement, all variables and given/known data
If the current is increasing at a rate of 0.1 A/sec what would the induced emf in a loop of 0.5 m be?


2. Relevant equations

[tex] I = \frac{|emf|}{R} [/tex]
[tex] emf = - \frac{d}{dt} B A cos(\theta) [/tex]


I'm obviously missing an equation or something.

Some guidance would be much appreciated.
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Mindscrape
Mindscrape is offline
#2
Dec2-07, 08:17 PM
P: 1,877
Where did you get your second equation from? That looks like something that you would see for a generator, and not a general formula.

So you know that
[tex]Emf = \epsilon = \frac{d \Phi}{dt}[/tex]

where the flux [itex]\Phi[/itex] would be

[tex]\Phi = \int B \cdot da[/tex]
mattst88
mattst88 is offline
#3
Dec2-07, 09:27 PM
P: 29
Thanks for your reply.

So multiply by dt

[tex] \epsilon dt = d \Phi [/tex]

Integrate both sides
[tex] \int \epsilon dt = \Phi [/tex]

Replace Phi

[tex] \int \epsilon dt = \int B \cdot dA [/tex]

B doesn't change with area, and is therefore constant. Integrate left side and replace A with area of a circle.

[tex] \int \epsilon dt = B \pi r^2 [/tex]

I don't quite know where to go from here.

Since Emf = IR, and we assume R is constant, if I is changing then E must be changing proportionally.

mattst88
mattst88 is offline
#4
Dec2-07, 09:32 PM
P: 29

Induced emf in a current loop


Is this valid?

[tex] \frac{d\epsilon}{dt} = \frac{dI}{dt} R[/tex]

[tex] \frac{d\epsilon}{dt} = 0.1 R [/tex]

[tex] d\epsilon = 0.1 R dt [/tex]

[tex] \epsilon = 0.1 R \int dt [/tex]

[tex] \epsilon = 0.1 R t[/tex]

And using that, plugging it into this equation:

[tex] \int \epsilon dt = B \pi r^2 [/tex]
[tex] 0.1 R \int t dt = B \pi r^2 [/tex]

Is this valid? Am I on the right track?
Mindscrape
Mindscrape is offline
#5
Dec3-07, 12:04 AM
P: 1,877
Very good! Almost there. I just realized I forgot to give you a minus sign too, sorry.

[tex]\epsilon = - \frac{d \Phi}{dt}[/tex]

One thing to consider is what the magnetic field of a ring actually is. This may be a confusing point, and it is, as you may say, "Wait a minute, I was told that biot-savart and all the others only works for magnetostatics." Well, that is very true, but we also need to calculate the magnetic fields. The only really good ways are with the magnetostatic methods. Basically this means that the magnetic field you calculate will only be an approximation, but the error is usually pretty small, unless you have very rapid fluctuations. We call this a quasistatic approximation.

Anyway, does it make sense that if you solve the magnetic field then you will be able to use that result to find an induced EMF? Since you will get a magnetic field with a current dependence, then the EMF will be the time derivative of your magnetic flux. In other words, you have flux [itex]\Phi = \pi r^2 B[/itex], find [itex]d \Phi/dt[/itex].

I'm surprised there isn't a part that tells you to calculate the induced electric field.


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