# Lenz's Law

1. Nov 7, 2009

### exitwound

This is not a homework problem due. It's practice. I have the answer of .198mV. I don't know how to get it.

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

2. Relevant equations

$$\phi_b = \int \vec B \cdot d\vec A$$

$$E = -\frac{d\phi_b}{dt}$$

3. The attempt at a solution

The magnetic field due to the solenoid is:

$$\mu_o i N$$
$$(1.26x10^{-6})(1.28 A)(85400 turns/m) = 1.37x10^{-1} T$$

The flux through the circular loop is:

$$\phi_b = \int \vec B \cdat d\vec A$$
$$\phi_b = BAcos 0 = BA$$

$$B = 1.37x10^{-1}$$ $$A=6.8x10{-3}$$
$$\phi_b = BAcos 0 = BA = (1.37x10^{1})(6.8x10^{-3})= 9.34x10^{-4} Wb$$

To find the EMF induced:
$$E = -\frac{d\phi_b}{dt}$$

I don't know where to go from here. How do I relate that 212rad/s to the problem?

2. Nov 7, 2009

### rock.freak667

Try putting I=I0sin(ωt) to get the magnetic field.

Then use Φ=NBA

and then use E=-dΦ/dt

3. Nov 7, 2009

### exitwound

$$B=\mu_o i_o N$$ we use:

$$B=\mu_o i_osin(\omega t)N$$ which leads to

$$\phi = BA = \mu_o i_osin(\omega t)NA$$

$$E = -\frac{d\phi_b}{dt}$$

$$E = -\frac{d}{dt}\mu_o i_osin(\omega t)NA$$

$$E = -\mu_o i_oNA\frac{d}{dt}(sin(\omega t))$$

Is this how you meant? I think I'm lost.

4. Nov 7, 2009

### rock.freak667

right yes and what is d/dt(sinωt) ?

5. Nov 7, 2009

### exitwound

As far as I can tell, (cos t)(ω)? Maybe??

6. Nov 7, 2009

### rock.freak667

So then

$$E=\mu_0 I_0 NBA \omega cos(\omega t)$$

so what is the amplitude?

7. Nov 7, 2009

### exitwound

The amplitude of the wave is 1.28 at maximum. But I dont know what that gets me.

8. Nov 7, 2009

### rock.freak667

no that is for the current

$$E=\mu_0 I_0 NBA \omega cos(\omega t)$$

What is the amplitude of E?

9. Nov 7, 2009

### exitwound

I don't understand.

10. Nov 7, 2009

### Seannation

The amplitude is equivalent to the maximum point on the wave that a wave equation describes.

Remember that cosine (and indeed sine) functions vary between -1 and 1, so the maximum cos value you can get is 1.

So what is the maximum that E can be in that equation?

11. Nov 7, 2009

### exitwound

(I think we have a typo. There shouldn't be a B in the equation, should there?)

Would it be: $$E=\mu_0 I_0 NA \omega$$
because cos (ωt)=1?

12. Nov 7, 2009

### rock.freak667

Yes the B should be there.

so the amplitude would be

$$E= \mu_0 I_0 NBA\omega$$

EDIT: sorry you are right, it is $E=\mu_0 I_0 NA \omega$