# EMF and magnetic flux

1. Apr 1, 2010

### gramentz

Here is how the question is stated exactly:

A circular conducting loop is held fixed in a uniform magnetic field thatt varies in time according to B(t) = B0exp(-at) where t is in s, a is in s-1 and B is the field strength in T at t=0. At t=0, the emf induced in the loop is 0.0758 V. At t= 1.76 s, the emf is 0.0192 V. Find a. Answer in units of s-1

2. Relevant equations
emf= $$\frac{d\Phi}{dt}$$

3. The attempt at a solution

I'm quite lost on this problem! I tried plugging in 0.0192 = B(t)/ 1.76 and solve for a, but obviously that did not produce the correct answer. Any help would be appreciated!

2. Apr 1, 2010

### gabbagabbahey

Hint: How is magnetic flux related to magnetic field?

3. Apr 1, 2010

### gramentz

The magnetic flux is a way of counting the total magnetic field lines. A changing magnetic field (given by the B(t)) will produce the emf's given.

4. Apr 1, 2010

### gabbagabbahey

No it isn't (thinking of it in this way allows you to visualize things, but the total number of field lines is infinite for all non-zero fields)

There is an equation that tells you how to determine the magnetic flux through a surface from the magnetic field (it involves an integral)...what is that equation?

5. Apr 1, 2010

### gramentz

$$\int$$BdA.......B(dot)dA = BAcos$$\theta$$

6. Apr 1, 2010

### gabbagabbahey

Right, when the magnetic field is uniform (such as in this case), the flux is given by $\Phi=\textbf{B}\cdot\textbf{A}=BA\cos\theta$

What is the surface area $A$ in this case? What does that tell you the EMF is?

7. Apr 1, 2010

### gramentz

I am not sure I understand because I don't see any values given in the problem for the dimensions of this loop. Do you mean A = $$\frac{\phi}{B}$$?

8. Apr 1, 2010

### gabbagabbahey

Sorry, you aren't given $A$, and you don't need to find it.

You have two equations, $\text{EMF}=-\frac{d\Phi}{dt}$ and $\Phi=BA\cos\theta$...put them together...what do you get?

9. Apr 1, 2010

### gabbagabbahey

Sorry, you aren't given $A$, and you don't need to find it.

You have two equations, $\text{EMF}=-\frac{d\Phi}{dt}$ and $\Phi=BA\cos\theta$...put them together...what do you get?

10. Apr 1, 2010

### gramentz

EMF= -$$\frac{B_{0}exp(-at)A}{dt}$$ ? Or take the derivative of BAcos$$\phi$$ with respect to time and set that equal to EMF? (Thank you for your help thus far though)

11. Apr 2, 2010

### gramentz

I'm thinking I have to use the emf at time t=0 to find the initial B at time t=0.

12. Apr 2, 2010

### gabbagabbahey

When you substitute your equation for $\Phi$ into the emf equation, you get

$$\text{EMF}=-\frac{d}{dt}\left(B_0e^{-at}A\cos\theta\right)$$

Do $A$, $\cos\theta$ or $B_0$ depend on time? Why or why not?

What does that make the derivative?

13. Apr 4, 2010

### gramentz

The only one that would be dependent on time would be B$$_{0}$$ because the loop is held fixed and the angle does not change.

14. Apr 4, 2010

### gabbagabbahey

$B_0$ is a constant. The only thing that varies with time is the exponential, so

$$\text{EMF}=-\frac{d}{dt}\left(B_0e^{-at}A\cos\theta\right)=-B_0A\cos\theta\frac{d}{dt}\left(e^{-at}\right)$$

right?

Carry out the derivative and then plug in the information you are given.