Finding the EMF Generated by a Solenoid

[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.

Homework Statement

Homework Equations

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

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

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?

 

[rock.freak667]

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

Then use Φ=NBA

and then use E=-dΦ/dt

 

[exitwound]

So instead of using

$$B=\mu_o i_o N$$ 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.

 

[rock.freak667]

So instead of using

$$B=\mu_o i_o N$$ 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.

right yes and what is d/dt(sinωt) ?

 

[exitwound]

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

 

[rock.freak667]

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

So then


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


so what is the amplitude?

 

[exitwound]

The amplitude of the wave is 1.28 at maximum. But I don't know what that gets me.

 

[rock.freak667]

The amplitude of the wave is 1.28 at maximum. But I don't know what that gets me.

no that is for the current



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

What is the amplitude of E?

 

[exitwound]

I don't understand.

 

[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?

 

[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?

 

[rock.freak667]

(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?

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 $E=\mu_0 I_0 NA \omega$