Induced Voltage of Rotating Circular Loop

[peroAlex]

HI! I'm a student, this question appeared in one of the old tests. I though I had it solved but apparently I must have missed somewhere. Could someone please at least hint me where I went wrong?

Homework Statement

Thin wire is wound into circular frame with radius $r = 0.05m$ which uniformly rotates around axis with $\omega = 45 rad/s$ in homogeneous magnetic field $B = 0.1 T$. Compute induced voltage $0.003 s$ after magnetic dipole of a loop is perpendicular to magnetic field lines.

Homework Equations

$U_i = B S \omega \sin{(\omega t)}$

The Attempt at a Solution

First I used upper equation (only the $B S \omega$ part) to compute amplitude of induced voltage which yields $U_0 \approx 0.0353 V$. Next I tried plugging $t = 0.003$ into full equation which returned $\approx 0.00476 V$. There are 4 options available (you chose the one which is correct), however, I surprisingly found out that this result is not among 4 possible answers.

Two "plausible" answers are $0.035 V$ (which is presumably the amplitude) and $0.0217 V$ which seems to be the correct result. The rest are $0.0637 V$ and $0.0438 V$, but they should be excluded because induced voltage cannot take values higher than amplitude, right?

Now, in order for #0.0217# to stand, $\sin{\omega t} = 0.63$ thus $\omega t = 0.68 rad$. This would occur when $t \approx 0.015s$ and not at given $0.003s$!

Remember how magnetic field and magnetic dipole of loop are perpendicular to each other? In this situation, induced voltage should be zero since flux is non-existent. How could it possibly increase to 60% of amplitude in such a short time span? I'm really lost here, could somebody please help me?

 

[cnh1995]

Remember how magnetic field and magnetic dipole of loop are perpendicular to each other? In this situation, induced voltage should be zero since flux is non-existent.

Zero flux does not mean zero induced voltage.
What is the relation between flux and induced emf?

Ui=BSωsin(ωt)Ui=BSωsin⁡(ωt) U_i = B S \omega \sin{(\omega t)}

What is the induced emf at t=0 in this equation?
What is the actual emf induced at t=0 in this problem?

 

[peroAlex]

OK, flux will equal zero when $\cos{\omega t} = 0$ this is when $t_0 = \frac{\pi}{2} s$. According to $U_i = - \frac{d (\text{flux)}}{dt} = - B S \omega \sin{(\omega t)}$ it means that at $t_0$ induced EMF will equal its full value (negative) $U_i = - B S \omega = - 0.035342917 V$. So if I add $t = t_0 + 0.003 s$ I get almost indistinguishable change yielding $U_i = - 0.035342757 V$. Am I on the right track? I still feel like something is missing...

 

[cnh1995]

this is when t0=π2st0=π2s t_0 = \frac{\pi}{2} s

No, π/2 is the angle, not time. At t=0, you have flux=0 (magnetic dipole of the coil is perpendicular to the external field).
So what is the equation for flux?
From that, what is the equation for induced emf?

 

[peroAlex]

Equation for flux is $\Phi = B S \cos{\omega t}$, but equation for induced emf is $U_i = - \frac{d \Phi}{dt} = B S \omega \sin{\omega t}$. I'm genuinly confused right now and I'm sorry for troubling you with that one, but if $t = 0s$ then $\cos{0} = 1$ and flux doesn't equal zero.

But if I want flux to be zero, the only way is to make $\cos{\omega t} = 0$, right? So if $45 \cdot t = \frac{\pi}{2}$, time has to equal $t = \frac{\pi}{90} \approx 0.03491 s$. If I plug this time into induced emf equation I get $U_i = 0.0353 V$.

I don't know what else I'm missing. I've searched through my lecture notes, been on the Internet... Still don't know what I'm missing.

 

[cnh1995]

Equation for flux is $\Phi = B S \cos{\omega t}$, but equation for induced emf is $U_i = - \frac{d \Phi}{dt} = B S \omega \sin{\omega t}$. I'm genuinly confused right now and I'm sorry for troubling you with that one, but if $t = 0s$ then $\cos{0} = 1$ and flux doesn't equal zero.

But if I want flux to be zero, the only way is to make $\cos{\omega t} = 0$, right? So if $45 \cdot t = \frac{\pi}{2}$, time has to equal $t = \frac{\pi}{90} \approx 0.03491 s$. If I plug this time into induced emf equation I get $U_i = 0.0353 V$.

I don't know what else I'm missing. I've searched through my lecture notes, been on the Internet... Still don't know what I'm missing.

At t=0, you have flux=0 (magnetic dipole of the coil is perpendicular to the external field).
So what is the equation for flux?

From this, you can write
Φ(t)=BAsin(ωt).
Compute dΦ/dt and put t=0.003s.

If I plug this time into induced emf equation I get Ui=0.0353VUi=0.0353V U_i = 0.0353 V.

I am getting the same answer upto three decimal places. Did you convert the angle from degree into radian?
What is the given answer?

 

[peroAlex]

Well, I did mention that at the beginning of the post,

Two "plausible" answers are $0.035 V$ (which is presumably the amplitude) and $0.0217 V$ which seems to be the correct result. The rest are $0.0637V$ and $0.0438V$, but they should be excluded because induced voltage cannot take values higher than amplitude, right?

But it seemed pretty counter-intuitive to think of $0.035V$ being the correct answer, simply because I probably assumed that just because flux is zero so is induced emf. So do you believe that this answer must be correct afterall?

 

[cnh1995]

So do you believe that this answer must be correct afterall?

Yes.

Assume the coil starts to rotate from zero-flux position i.e. at t=0, the flux is zero.
Therefore,

Φ(t)=BAsin(ωt).

Its derivative is a cosine function, so at t=0.003s, you get induced voltage of 0.035V.
(You should convert set your calculator in radian mode, or you'll get V=0.03534 V.
There isn't much difference between the two, but this is simply because the angle is very small and cosine of a small angle is close to 1. It won't hold for larger angles.)

Also, this voltage is very close to the amplitude for the same reason. The angle is so small that its cosine is very close to 1.
You can see the difference after three decimal places.

 

[peroAlex]

Thank you so much!
I was wondering how andwhy, but now I finally understand it. Thank you so much for helping me on that one ;)