Find k from magnetic field and magnetic flux.

In summary, the student is trying to solve for k in order to calculate the magnetic flux through a circuit, but doesn't understand why the field isn't uniform over the surface.
  • #1
Ylle
79
0

Homework Statement


Hello...

I got a problem I really can't figure out.
I have the scenario in this link: http://www.gratisupload.dk/download/41677/"

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:

[tex]B=ky{{e}^{-{{t}^{2}}/{{\tau }^{2}}}}{{e}_{z}}$[/tex]
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:

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

Now determine k

Homework Equations



[tex]\[{{\Phi }_{B}}=BA\][/tex]

The Attempt at a Solution



I know the answer is supposed to be:

[tex]\[k=2{{B}_{0}}/L\][/tex]

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.
 
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  • #2
Ylle said:

Homework Equations



[tex]\[{{\Phi }_{B}}=BA\][/tex]

This formula is only true if [itex]\textbf{B}[/itex] is uniform over the surface and normal to the surface. Are these two conditions met by the [itex]\textbf{B}[/itex] 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
Ahhh, I guess, since the field is in z-direction it's not uniform.
So what I need to do is:

[tex]\int[/tex][tex]\int B dx dy[/tex] with the limits 0 to L in both integrals, and the equal the flux I have, and solve for k ?
 
  • #4
Ylle said:
Ahhh, I guess, since the field is in z-direction it's not uniform.

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

So what I need to do is:

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

Yup.
 
  • #5
I see :)

Thank you very much.
 

1. How do you find the value of k from magnetic field and magnetic flux?

The value of k can be found by dividing the magnetic flux (Φ) by the magnetic field (B), as k = Φ/B. This equation is derived from the definition of magnetic flux, which is the product of magnetic field and the area that it passes through.

2. What are the units for k in this equation?

K has the units of webers per ampere (Wb/A) or henrys (H). These units are derived from the units of magnetic flux (webers) and magnetic field (teslas).

3. Is there a specific direction for k in this equation?

Yes, the direction of k is determined by the direction of the magnetic field and the direction of the area in which it passes through. The direction of k is perpendicular to both the magnetic field and the area.

4. Can this equation be used for any type of magnetic field?

Yes, this equation can be used for any type of magnetic field, whether it is a uniform or non-uniform field. However, in the case of a non-uniform field, the value of k may vary at different points in the field.

5. What is the significance of finding the value of k in this equation?

Finding the value of k allows us to quantify the strength of the magnetic field in a given area. It also allows us to calculate other properties, such as the magnetic force on a charged particle moving through the field.

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