Magnitude and Direction of Induced Current on a wire around a cylindrical volume

[Aphillips2010]

Homework Statement

A uniform magnetic field B is confined to a cylindrical volume of radius 0.080m. B is directed into the plane of the paper and is increasing at a constant rate of (delta)B/(delta)t=0.300 T/s. Calculate the magnitude and direction of the current induced in a circular wire ring of radius 0.16m and resistance 1.4 Ohms that encircles the magnetic field region.

Homework Equations

The Attempt at a Solution

step1: find emf through wire ring using magnetic flux of cylinder?
step2: R(resistance)=1.4Ohms, use equation I=emf/R to find magnitude. and direction is going to be counter clockwise to appose change in flux.

Not sure how to go about starting on this problem. Help please.

 

[dikmikkel]

You could start by finding the flux phi:
[itex] \Phi = \int \vec{B}\cdot \text{d}\vec{A}[/itex]
But already you know how much the flux changes pr. time so:
[itex] \varepsilon = -\dfrac{\text{d}\Phi}{\text{d}t} = -\dfrac{\text{d}}{\text{d}t}\int \vec{B}\cdot \text{d}\vec{A} = [/itex]
? You know that the area dosen't change with time and you could switch the order of integration and differentiation?

 

[Aphillips2010]

Do i already know how much the flux changes or how much the magnitude of the magnetic field changes?

I keep confusing myself and I don't even know where to start anymore

 

[dikmikkel]

You know dB/dt and that the magnetic field is uniform, and that you can change the order of diff and int cause the integral is time-independent. So you know how much B is changing and how much the flux canges because you also know your area(i assume that the B-field is parallel with the z axis in the cylindrer)

 

[Aphillips2010]

yes, my Area vector is also parallel and in the same direction of B.

 

[Aphillips2010]

ε=(dB/dt)(∏r^2)cos(0°) is that looing right? and I am unsure about what to do with the different radius measurements too.

 

[dikmikkel]

Okay i can't see why you want to do something about the radius but except for that you divide the whole thing by R and assume quasistationary current so it is okay to use: emf = RI
So what is the problem, you know all the things, just plug in.
dB/dt = 0.300T/s, A = Pi x 0.080^2m^2.
There is only 1 radius, the cylinder's radius
Also there is a sign error i get as a final result:
[itex] I = \dfrac{\varepsilon}{R} = -\dfrac{\dfrac{\text{d}B}{\text{d}t} \pi r^2}{R}[/itex]
You are aware that cos(0deg) = 1 right?
As a final remark: Use Lentz' Law to find the direction of current.

 

[Aphillips2010]

so the radius of the wire circling the cylinder plays no part in the equation? and yes i am aware haha. the negative current in the end identifies the direction as being counterclockwise, am i correct?

 

[dikmikkel]

When you use Ampere's Law you make up an imaginary loop and the radius is only relevant for [itex]\int d\vec{A}[/itex]
It is an approximation though.
That is a bigger loop gives bigger current.
Yes i believe you are correct: The nature will always try to oppose a change in magnetic flux hence current runs counterclockwise as indicated on the sign of the EMF.