# Homework Help: Find k from magnetic field and magnetic flux.

1. Mar 22, 2010

### Ylle

1. The problem statement, all variables and given/known data
Hello...

I got a problem I really can't figure out.

Besides that I know that there is a magnetic field that works everywhere and is along the z-axis. This field, that depends on the y-coordinate and the time t is given as:

$$B=ky{{e}^{-{{t}^{2}}/{{\tau }^{2}}}}{{e}_{z}}$$
where tau is a positive time-constant, k is a constant with dimension T/m and ez is unit vector in the direction of the z-axis.

The magnetic field raises a magnetic flux through the circuit given by:

$$${{\Phi }_{B}}={{B}_{0}}{{L}^{2}}{{e}^{-{{t}^{2}}/{{\tau }^{2}}}}$$$
where B0 is a positive constant with dimension T.

Now determine k

2. Relevant equations

$$${{\Phi }_{B}}=BA$$$

3. The attempt at a solution

I know the answer is supposed to be:

$$$k=2{{B}_{0}}/L$$$

It seemed to good to be true if I just inserted the flux and the magnetic field into this equation, and then setting A = L2.

If I did that I got: k = B0 / y.

And I've been searching my book for examples and stuff I could use. But I can't come up with anything when I only have the magnetic field and magnetic flux. So I'm thinking there must be a trick that I'm not aware of :S

So can anyone point me in the right direction?

Regards.

Last edited by a moderator: May 4, 2017
2. Mar 22, 2010

### gabbagabbahey

This formula is only true if $\textbf{B}$ is uniform over the surface and normal to the surface. Are these two conditions met by the $\textbf{B}$ in your problem and the square surface bounded by the circuit you are given?

If not, you will need to use the more general definition of magnetic flux.

3. Mar 22, 2010

### Ylle

Ahhh, I guess, since the field is in z-direction it's not uniform.
So what I need to do is:

$$\int$$$$\int B dx dy$$ with the limits 0 to L in both integrals, and the equal the flux I have, and solve for k ?

4. Mar 22, 2010

### gabbagabbahey

The reason the field isn't uniform over the surface, is because it depends on $y$ and $y$ varies over the surface.

Yup.

5. Mar 22, 2010

### Ylle

I see :)

Thank you very much.