Calculating Energy Dissipation in a Rectangular Coil with Time-Varying Current

[arl146]

Homework Statement

A rectangular coil with N = 2,000 turns that has a resistance of 7.90 Ohms is coplanar with a long wire which carries a current which depends on time according to I0 *exp(-t/tau), where I0 = 6.50 A and tau = 4.30 s. The rectangular loop has a width of W = 2.00 cm and length L = 7.60 cm. The near side of the loop is a distance D = 2.90 cm from the wire.

What is the total energy dissipated in the entire coil from t = 0 to t = 2.60 s?

Homework Equations

I=I0*exp(-t/tau)
Flux= [L*mu*I*ln(1+w/d)] / 2*pi
V (or EMF) = d(flux)/dt = [ (L*mu*ln(1+w/d)) / 2*pi ]*(-I0*exp(-t/tau))
where this, (-I0*exp(-t/tau)) , is the time derivative of I
P=(V^2)/R
Energy= integral(P)

The Attempt at a Solution

I already got the previous questions for this problem. The answers resulted in: Flux for one turn of the coil at 2.6 seconds= 2.83×10-8 T*m^2, EMF for entire coil at 2.6 seconds= 1.32×10-5 V, and the power dissipated from the coil at 2.6 seconds=2.19×10-11 W.

I just don't know how to integrate the power equation.
Thanks!

 

[ehild]

You left out 1/tau while differentiating with respect to time.
The dissipated energy is the integral of the power from t= to t=2.6 s. And the power is (V^2)/R as you have shown. The emf is of the form C exp(-t/tau) where C is a constant. What is its square?
You certainly know what is the integral of an exponential function.

ehild

 

[arl146]

You left out 1/tau while differentiating with respect to time.
The dissipated energy is the integral of the power from t= to t=2.6 s. And the power is (V^2)/R as you have shown. The emf is of the form C exp(-t/tau) where C is a constant. What is its square?
You certainly know what is the integral of an exponential function.

ehild

Well, I actually had the 1/tau in my work I just accidentally didn't type it. Well, the intergral of exp(-t/tau) would just be -tau*exp(-t/tau) ?? So the emf would just be -(tau^2)*exp(-(t^2)/(tau^2)) ? and divide by same R i was using ? It's that simple ? What would the constant be?

 

[ehild]

Take care when you square an exponential function. (exp(x))^2=exp(2x)
And the square of a negative quantity is positive. ehild

 

[arl146]

Take care when you square an exponential function. (exp(x))^2=exp(2x)
And the square of a negative quantity is positive.


ehild

Wait, I'm confused so you square it first then take the integral or what

 

[ehild]

First square the emf as P=(V^2)/R , then integrate with respect to the time.

You wrote in a previous post that

"V (or EMF) = d(flux)/dt = [ (L*mu*ln(1+w/d)) / 2*pi ]*(-I₀*exp(-t/tau))
where this, (-I₀*exp(-t/tau)) , is the time derivative of I"

Here is a 1/tau missing. The time derivative of I is -I₀*1/(tau)*exp(-t/tau), so the emf is

V=[ (L*mu*ln(1+w/d)) / 2*pi ]*(-I₀(1/tau)*exp(-t/tau)).

You have to square this whole thing, divide by R and integrate.

ehild

 

[arl146]

First square the emf as P=(V^2)/R , then integrate with respect to the time.

You wrote in a previous post that

"V (or EMF) = d(flux)/dt = [ (L*mu*ln(1+w/d)) / 2*pi ]*(-I₀*exp(-t/tau))
where this, (-I₀*exp(-t/tau)) , is the time derivative of I"

Here is a 1/tau missing. The time derivative of I is -I₀*1/(tau)*exp(-t/tau), so the emf is

V=[ (L*mu*ln(1+w/d)) / 2*pi ]*(-I₀(1/tau)*exp(-t/tau)).

You have to square this whole thing, divide by R and integrate.

ehild

Yea, I know I left that out by accident. So V^2= [L^2*mu^2*I0^2*(ln(1+w/d))^2*exp(-2t/tau)] / 4*pi^2*tau^2 ?

So P= V^2/R= [(L^2*mu^2*I0^2*(ln(1+w/d))^2*exp(-2t/tau)) / 4*pi^2*tau^2] / R

Take the integral of that, would you pull everything out except for exp(-2t/tau) ? So really you are just integrating that, the rest a constants ? So the integral of exp(-2t/tau) would equal -((tau*exp(-2t/tau))/2) ?



Or no, it wouldn't be exp(-2t/tau) it would be exp(t^2/tau^2) ? but to integrate that I'm not sure of .. Ughh, I'm so confused for no reason =(

 

[ehild]

So P= V^2/R= [(L^2*mu^2*I0^2*(ln(1+w/d))^2*exp(-2t/tau)) / 4*pi^2*tau^2] / R

Take the integral of that, would you pull everything out except for exp(-2t/tau) ? So really you are just integrating that, the rest a constants ? So the integral of exp(-2t/tau) would equal -((tau*exp(-2t/tau))/2) ?

Yes, integrate the exponential function between t=0 and t=2.6 s and multiply by everything else.

Or no, it wouldn't be exp(-2t/tau) it would be exp(t^2/tau^2) ? but to integrate that I'm not sure of .. Ughh, I'm so confused for no reason =(

Forget exp(t^2/tau^2). exp(x) is e on the x-th power e^x. (e^x)^2=(e^x)*(e^x)=e^(x+x)=e^2x.

ehild

 

[arl146]

Forget exp(t^2/tau^2). exp(x) is e on the x-th power e^x. (e^x)^2=(e^x)*(e^x)=e^(x+x)=e^2x.

ehild

Oh, okay. Thanks ^.^ I'll try it out and if there's a problem, I'll be back after I try it out a few times! Hopefully I won't have to.

 

[arl146]

Okay, I have a problem. This is all of my work:


V= -\frac{L \mu ln(1+w/d) (I0) exp(-t/\tau)}{2 \pi \tau}

V² = \frac{L^2 \mu^2 (ln(1+w/d))^2 I0^2 exp(-2t/\tau)}{4 \pi^2 \tau^2}


P= \frac{V^2}{R}

\int P dt = Energy = \frac {L^2 \mu^2 (I0^2) (ln(1+w/d))^2}{4 \pi^2 \tau^2 R} * \int exp(-2t/\tau) dt

E= \frac {L^2 \mu^2 (I0^2) (ln(1+w/d))^2}{4 \pi^2 \tau^2 R^2} * [ exp(-2t/tau)*(-tau/2) ]

the boundaries on that is 0 to 2.6 ;
my answer i got after many times of trying was 2.773358998E-17 J
What is wrong ?

 

[ehild]

I do not see N (the 2000 turns).

ehild

 

[arl146]

Ohhhhh ... *facepalm* I even thought of that but figured that wasn't right so I didn't even try it. Thanks! again!

 

[arl146]

Can I just multiply by answer above by N^2 ?

 

[ehild]

Can I just multiply by answer above by N^2 ?

It would be enough, but there is some other computational error somewhere. It was not a good idea to square everything separately. Evaluate the constant of V first before squaring. And you can cancel pi as mu=4pi˙10^-7.

ehild

 

[arl146]

I mean, how do you square it without squaring each thing separately ? ...and what do you mean evaluate the constant of V first?

 

[ehild]

Here is a simple example: evaluate (2*3)². You can do it by evaluating 2*3 first, then squaring the result: it is 6²=36,
or squaring separately: 2²*3²=4*9=36.

And I found just now that you divided V^2 by R^2 instead of R.

ehild

 

[arl146]

No, I divided by R I don't know why I put R^2. But I got the answer. All I needed to do was multiply by N^2. thanks for the help

 

[ehild]

Splendid!
You are welcome.

ehild