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: <br /> E=\mu_0 I_0 NA \omega<br />
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: <br /> E=\mu_0 I_0 NA \omega<br />
because cos (ωt)=1?

Yes the B should be there.

so the amplitude would be

E= \mu_0 I_0 NBA\omegaEDIT: sorry you are right, it is E=\mu_0 I_0 NA \omega