Square coil rotated in a B field

[arl146]

Homework Statement

A 40 turn square coil has an area 11.0 cm2 and is rotated with an angular speed of 110.0 rads in a uniform 1.00 T magnetic field. The axis of the coil is perpendicular to the field at all times.

(i.) What is the frequency of the sinusoidal voltage induced across its ends?
(ii.) What is the peak voltage induced across the ends of the coil?
(iii.) What is the flux through the coil when the plane of the coil is inclined at 43.0° to the magnetic field?
(iv.) What is the average power dissipated in the resistor?

Homework Equations

The Attempt at a Solution

I got (i.) which was 17.5; There was also another question relating to "If the coil forms a closed circuit with a series resistance of 50.0 , what is the instantaneous power being dissipated in the resistor when the plane of the coil makes an angle of 43.0° with the magnetic field? " He gave the answer which was 0.469 W. I am not sure if that's useful for anything. A student gave an equation for part 2 -> Flux=BAcos(wt) which then you take the derivate and then either take the derivate again or say that the max of cos or sin is always 1 and use that. But I don't know how to take the derivate of that formula and then what to do next? Help please!

 

[yoshtov]

The equation for magnetic flux is only part of the story. Notice that this is an "Induced EMF" problem.

Other equations of interest are:

EMF=(-N)*(dFlux/dt)

where "N" is the total number of loops, and (dFlux/dt) is the derivative of flux w.r.t. time.

You are correct in stating that you need to take the derivative of the flux equation (in order to know what (dFlux/dt) is). However, this is not a difficult task at all! In the flux equation, (B) is constant, (A) is constant, the only variable that is actually changing with time is the angle that the coil makes with the B field, i.e., angular velocity.

So, (dFlux/dt) = B*A*(dcos(w)dt).

You still need to differentiate (dcos(w))/dt, but I think you can do that. It looks like your fellow student tried to do that, but didn't correctly identify the differentiated equation.

Part III is easy, it's just plug-n-chug, using the flux equation, Flux=B*A*cos(w)

Part IV: Remember that ALL of the energy in this circuit is dissipated through the resistor.

 

[arl146]

Okay, I'm still confused. So, for part 2 I use EMF=(-N)(dFlux/dt) so I don't understand what to do with (dFlux/dt) = B*A*(dcos(w)/dt), wouldn't (dcos(w)/dt) just be cos(1) then? And part 3 I just plugged the numbers in and it didn't work, they give me an angle how do I use it with that equation? For Part 4, the only equation I have for power is P=IV=(I^2)R=(V^2)/R would I use the last one, but what would V be?

 

[arl146]

Or no, (dcos(w)/dt) would just be sin(w) ? Still confused about the rest though.

 

[arl146]

still need help please !

 

[arl146]

How do you differentiate dcos(w)/dt ??

 

[yoshtov]

(I). I got 17.5 Hz as well.

(II). Since $$\epsilon =-N\frac{d\Phi}{dt}$$ and $$\Phi =BA{cos\boldsymbol{\theta}}$$, and the derivative of that is $$\frac{d\Phi }{dt}=-BA{sin}\theta \frac{d\theta }{dt}$$, we can say that the highest EMF value will be present when dFlux/dt is highest, and THAT will be the highest when sin is at a maximum, when it equals 1. So, plugging in all of our known values, we get $$\epsilon _{max}=NBA\frac{d\theta }{dt}=40\times 1\times 0.0011\times 110=4.84 V$$.

(III). Remember that, when calculating flux, you want to use the normal of the coil. In this case, the 43° is referring to the plane of the coil, and not the normal of the plane of the coil. Subtract 43° from 90° to get 47°, and use that in the flux equation to get 0.075 Wb of flux.

(IV). It's asking for average power dissipated, so we have to calculate the RMS voltage of the induced EMF. We have the maximum EMF, and we can get the RMS value using this equation.

$$X_{rms}=\frac{X_{max}}{\sqrt{2}}=\frac{4.84}{\sqrt{2}}=3.42V$$

and then

$$P=\frac{V^{2}}{R}=\frac{3.42^{2}}{50}=0.23 W$$ I'm not sure why my answer differs from the answer that you have for this problem. I do notice that my answer is exactly 1/2 of the one you have.

(IV-2). Looking back at the three equations above, we can calculate the instantaneous EMF present when the coil is at 43° (when the normal is at 47°). $$\epsilon =NBA{sin}\theta \frac{d\theta }{dt}=40\times 1\times 0.0011\times {sin}47\times 110=3.54 V$$

Remember that $$P=IV=I^{2}R=\frac{V^{2}}{R}$$

So, we have the voltage (EMF) and we have the resistance, so we can calculate instantaneous power dissipation as follows:

$$P=\frac{V^{2}}{R}=\frac{3.54^{2}}{50}=0.25 W$$

I hope this helps!

 

[arl146]

Ohh, I'm so stupid that all makes so much sense. Thank you, especially for taking the time out to make it all easier to read! I had my professor reset my tries on the first 2 but I got the last one right.

 

[arl146]

Well, no part 3 was wrong.

 

[yoshtov]

Sadly, we all make stupid mistakes, and I am no exception. Recall that flux is $$\Phi =BAcos(\theta )$$ Remember that area is NOT 0.11 m^2, but rather, 0.0011 m^2, as there are (100^2) square-centimeters in a square-meter. Anyway, run with the "adjusted" area, and find that $$\Phi =1\times 0.0011\times cos(47)=0.00075=7.5\times 10^{-4}$$Wb of flux.

 

[arl146]

yea i did that. it didn't work out =/

 

[arl146]

oh i got it. they wanted you to take into consideration the number of turns for some reason.. i don't know why since flux you don't usually deal with the number of turns unless you want EMF

 

[yoshtov]

i spoke with my engineering professor, and he said that there is a significant difference between "flux through a loop of wire" and "flux through a coil." He also said that this is something physics teachers argue about. In short: $$\Phi _{loop}=B\times A\times cos(\theta )$$ and $$\Phi _{coil}=N\times B\times A\times cos(\theta )$$ where "N" is number of loops. I am sorry I didn't acknowledge the number of loops in my calculations.

 

[arl146]

Oh, that's okay! I still got it right in the end, mostly all because of you. But that's weird, my book didn't show it like that. It just showed the first one. Thanks!