# Induced emf: Faraday's Law and Lenz's Law Question

1. Feb 5, 2008

### predentalgirl1

[SOLVED] Induced emf: Faraday's Law and Lenz's Law Question

The plane of a conductive loop with an area of 0.020 m^2 is perpendicular to a uniform magnetic field of 0.30 T. If the field drops to zero in 0.0045 s, what is the magnitude of the average emf induced in the loop?

Given that,
Area (a) = 0.020m2
Magnetic filed (B) = 0.30 T.
When the field drops to zero,
Time taken = 0.0045 s
I have,
Induced emf = Change in flux
Time taken
= B x a cos 90
Time taken
= 0.30 x 0.020 x 1
0.0045
= 1.33 V

2. Feb 5, 2008

### rock.freak667

That should be correct.

3. Feb 5, 2008

### predentalgirl1

Is it really correct...or are you just telling me that?

4. Feb 5, 2008

### rock.freak667

I calculated it the same way as you and got the same answer

$$E=\frac{BA}{t}$$

5. Feb 5, 2008

### predentalgirl1

Ok...thanks. I just can't believe that I did it right. :)

6. Feb 5, 2008

### Mindscrape

$$\Phi = \iint \mathbf{B} \cdot d\mathbf{a}$$

which in a constant magnetic field perpindicular to a constant area will reduce to

$$\Phi = BA$$

Then you know that

$$\epsilon = \frac{d \Phi}{dt}$$

So the emf would be (assuming a constant change in the magnetic field)

$$\epsilon = \frac{\Delta \Phi}{\Delta t}$$

and further shows that

$$\epsilon = \frac{A * (B_i - B_f)}{(t_i - t_f)}$$

where B_f and t_f are zero.

With numbers

$$\epsilon = \frac{.02*.3}{.0045}\frac{mB}{s^2} = 1.33V$$

You need some more confidence in yourself.

Edit:
Oh right, what about Lenz's law?

Last edited: Feb 5, 2008
7. Feb 5, 2008

### predentalgirl1

When it comes to physics I get so confused, that's why I'm on here. lol But thanks for your help.