Conductive loop that is contracting in a magnetic field

[adrms]

Homework Statement

A conductive loop is painted around a plastic balloon. A magnetic field B = 0.2 cos(4t)[T] is applied perpendicular to the equatorial plane. The balloon is contracting with a radial velocity $v$. When the radius of the balloon is 0.5[m], the induced RMS voltage in the loop is 500[mV].
Find the velocity v of radial contraction in that moment.

Relevant Equations

$$V = \int \left(\vec{v} \times \vec{B}\right) \bullet \vec{dl} - \int _S \frac{\vec{dB}}{dt} \bullet \vec{ds}$$
$$V_{\text{rms}} = \frac{V_{\text{max}}}{\sqrt{2}}$$

$$V = \int \left(\vec{v} \times \vec{B}\right) \bullet \vec{dl} - \int _S \frac{\vec{dB}}{dt} \bullet \vec{ds}$$
From the statement I know that: B⊥v, (B x v) // dl and B // ds.
$$V = \int vBdl - \oint _S \frac{dB}{dt} ds$$

v is the speed with which all the segments dl are aproximating to the center of the loop. It is constant and does not depend on the length of the circunsference so it can be separated from the first integrand.

$$V = v \int B dl - \oint _S \frac{dB}{dt} ds$$

Can B be considered constant in the moment when radius of the loop is r=0.5?
Can I do the same with $\frac{dB}{dt}$ ?

In that case I will have
$$V = vB \int dl - \frac{dB}{dt} \oint _S ds$$
$$V = vB (2 \pi r) - \frac{dB}{dt} (\pi r ^2)$$
$$v = \frac{V + \frac{dB}{dt} (\pi r ^2)}{B (2 \pi r)}$$
If I replace the variables
$$v = \frac{ 500 \sqrt{2} \times 10^{-3} - 0.2 \pi \sin(4t)}{0.2 \pi \cos (4t)}$$
But I do not know the value of t so I do not know how to reach a numeric result.

---

The problem is from "Electromagnetics with Applications" 5th Edition by John Daniel Kraus

 

[scottdave]

Do you know how sinusoidal voltage signals relate to the RMS voltage?

 

[collinsmark]

Hello @adrms,

Welcome to PF!

Problem Statement: A conductive loop is painted around a plastic balloon. A magnetic field B = 0.2 cos(4t)[T] is applied perpendicular to the equatorial plane. The balloon is contracting with a radial velocity $v$. When the radius of the balloon is 0.5[m], the induced RMS voltage in the loop is 500[mV].
Find the velocity v of radial contraction in that moment.
Relevant Equations: $$V = \int \left(\vec{v} \times \vec{B}\right) \bullet \vec{dl} - \int _S \frac{\vec{dB}}{dt} \bullet \vec{ds}$$
$$V_{\text{rms}} = \frac{V_{\text{max}}}{\sqrt{2}}$$

$$V = \int \left(\vec{v} \times \vec{B}\right) \bullet \vec{dl} - \int _S \frac{\vec{dB}}{dt} \bullet \vec{ds}$$
From the statement I know that: B⊥v, (B x v) // dl and B // ds.
$$V = \int vBdl - \oint _S \frac{dB}{dt} ds$$

v is the speed with which all the segments dl are aproximating to the center of the loop. It is constant and does not depend on the length of the circunsference so it can be separated from the first integrand.
View attachment 244297

$$V = v \int B dl - \oint _S \frac{dB}{dt} ds$$

Can B be considered constant in the moment when radius of the loop is r=0.5?
Can I do the same with $\frac{dB}{dt}$ ?

In that case I will have
$$V = vB \int dl - \frac{dB}{dt} \oint _S ds$$
$$V = vB (2 \pi r) - \frac{dB}{dt} (\pi r ^2)$$
$$v = \frac{V + \frac{dB}{dt} (\pi r ^2)}{B (2 \pi r)}$$
If I replace the variables
$$v = \frac{ 500 \sqrt{2} \times 10^{-3} - 0.2 \pi \sin(4t)}{0.2 \pi \cos (4t)}$$
But I do not know the value of t so I do not know how to reach a numeric result.

---

The problem is from "Electromagnetics with Applications" 5th Edition by John Daniel Kraus

The problem poorly constructed. The main reason (and there may be a couple of reasons, btw) is that the concept of "RMS" is ill defined for a non-stationary signal. This signal is not stationary. While the signal has sinusoidal properties (sinusoidal signals are stationary), this signal is not sinusoidal. It's a damped sinusoidal signal and those are not stationary.

You can still measure the RMS voltage of a non-stationary signal if you have a precisely defined measurement interval. And then the answer would only apply for that particular interval. However, that's not how this problem was worded.

It's not your fault, @adrms, but shame on the author of your coursework.

---

Anyhoo, you can still "sort of" do this problem by assuming that the RMS signal is the "peak" of any sinusoidal-like signal divided by $\sqrt{2}$. The assumption isn't technically true here, but it will lead you to an answer.

Your approach seems valid through your

$v = \frac{V + (\pi r ^2) \frac{dB}{dt}}{B (2 \pi r)}$

equation. So far so good. (I would have done it differently by simply invoking Faraday's law, $\varepsilon = - \frac{d \{BA \}}{dt}$ and use the product rule when differentiating, since both $B$ and $A$ are functions of time. But your approach gives essentially the same equation. [There might be a difference in a minus sign, but that's a difference in how you interpret the physical significance of each term.])

But something went wrong here:

$v = \frac{ 500 \sqrt{2} \times 10^{-3} - \color{red} {0.2 \pi \sin(4t)}}{0.2 \pi \cos (4t)}$

You need to use the chain rule when taking the derivative. It's not just switching sines and cosines. You need to pay attention to "what's inside" the sines and cosines too. [Edit: Oh, and where did the $r^2$ go? And if you haven't applied the $V_{RMS} = \frac{V_{\rm{peak}}}{\sqrt{2}}$ yet, why is there a $\sqrt{2}$ attached to the 0.5 V?]

And in the denomintor, you forgot a 2 and an $r$.

Once you have that fixed up, you can get rid of the "$t$" by using the not-so-true-here assumption/approximation, where $V_{RMS} = \frac{V_{\rm{peak}}}{\sqrt{2}}$, which we discussed above.

 

[adrms]

Hello @adrms,

Welcome to PF! The problem poorly constructed. The main reason (and there may be a couple of reasons, btw) is that the concept of "RMS" is ill defined for a non-stationary signal. This signal is not stationary. While the signal has sinusoidal properties (sinusoidal signals are stationary), this signal is not sinusoidal. It's a damped sinusoidal signal and those are not stationary.

You can still measure the RMS voltage of a non-stationary signal if you have a precisely defined measurement interval. And then the answer would only apply for that particular interval. However, that's not how this problem was worded.

It's not your fault, @adrms, but shame on the author of your coursework.

---

Anyhoo, you can still "sort of" do this problem by assuming that the RMS signal is the "peak" of any sinusoidal-like signal divided by $\sqrt{2}$. The assumption isn't technically true here, but it will lead you to an answer.

Your approach seems valid through your

$v = \frac{V + (\pi r ^2) \frac{dB}{dt}}{B (2 \pi r)}$

equation. So far so good. (I would have done it differently by simply invoking Faraday's law, $\varepsilon = - \frac{d \{BA \}}{dt}$ and use the product rule when differentiating, since both $B$ and $A$ are functions of time. But your approach gives essentially the same equation. [There might be a difference in a minus sign, but that's a difference in how you interpret the physical significance of each term.])

But something went wrong here:

$v = \frac{ 500 \sqrt{2} \times 10^{-3} - \color{red} {0.2 \pi \sin(4t)}}{0.2 \pi \cos (4t)}$

You need to use the chain rule when taking the derivative. It's not just switching sines and cosines. You need to pay attention to "what's inside" the sines and cosines too. [Edit: Oh, and where did the $r^2$ go? And if you haven't applied the $V_{RMS} = \frac{V_{\rm{peak}}}{\sqrt{2}}$ yet, why is there a $\sqrt{2}$ attached to the 0.5 V?]

And in the denomintor, you forgot a 2 and an $r$.

Once you have that fixed up, you can get rid of the "$t$" by using the not-so-true-here assumption/approximation, where $V_{RMS} = \frac{V_{\rm{peak}}}{\sqrt{2}}$, which we discussed above.

First of all, thank you for welcoming me to the forum

The problem poorly constructed. The main reason (and there may be a couple of reasons, btw) is that the concept of "RMS" is ill defined for a non-stationary signal. This signal is not stationary. While the signal has sinusoidal properties (sinusoidal signals are stationary), this signal is not sinusoidal. It's a damped sinusoidal signal and those are not stationary.

You can still measure the RMS voltage of a non-stationary signal if you have a precisely defined measurement interval. And then the answer would only apply for that particular interval. However, that's not how this problem was worded.

I was not aware of that fact. Thank you for clarifying it.

But something went wrong here:

$v = \frac{ 500 \sqrt{2} \times 10^{-3} - \color{red} {0.2 \pi \sin(4t)}}{0.2 \pi \cos (4t)}$

I'm sorry, I think I may have messed it up while taking the derivative and replacing the variables.

Please let me start with this equation:

$$v = \frac{V + (\pi r ^2) \frac{dB}{dt}}{B (2 \pi r)}$$

So I need to take the derivative of $B$ with respect to $t$.

$$\frac{dB}{dt} = - 0.2 \sin (4t) (4)$$
$$\frac{dB}{dt} = - 0.8 \sin (4t)$$

Then

$$v = \frac{V - 0.8 \sin(4t) (\pi r^2)}{B (2 \pi r)}$$

Replacing with the values of the statement (r=0.5m)
$$v = \frac{V - 0.8 \sin(4t) (\pi 0.5^2)}{(0.2 \cos(4t)) (2 \pi 0.5)}$$
$$v = \frac{V - 0.2 \pi \sin(4t)}{0.2 \pi \cos (4t)}$$

From the statement I know that when radius is 0.5m, the RMS induced voltage is 0.5V, so I can do:

$$V_{\text{RMS}} = \frac{V_{\text{peak}}}{\sqrt{2}}$$
$$V_{\text{peak}} = \sqrt{2} V_{\text{RMS}}$$

Can I just replace in $V$?
$$v = \frac{ 0.5 \sqrt{2} - 0.2 \pi \sin(4t)}{0.2 \pi \cos (4t)}$$

I need the parameter $t$ to correspond for when $V$ takes its peak value. But I just do not see the relation... I do not see how can I get rid of the $t$.

I have been playing with the graphs of the situation and I found that if I want $V_{\text{peak}}$ to be equals to $\sqrt{2} V_{\text{RMS}} = 0.5 \sqrt{2}$ I need the velocity $v$ to be approximately equals to 0.51625 and in consecuence $t=0.2736$

The green curve is the curve that correspond to the voltage $V = vB (2 \pi r) - \frac{dB}{dt} (\pi r ^2)$. Desmos Link.

But without the graph and the approximations I am lost with the equation

$$v = \frac{ 0.5 \sqrt{2} - 0.2 \pi \sin(4t)}{0.2 \pi \cos (4t)}$$

I tought that maybe I could find $t$ by finding what value of $t$ verifies $\frac{dV}{dt} = 0$. But for that I would need the value of velocity $v$...

 

[collinsmark]

First of all, thank you for welcoming me to the forum
I was not aware of that fact. Thank you for clarifying it.
I'm sorry, I think I may have messed it up while taking the derivative and replacing the variables.

Please let me start with this equation:

$$v = \frac{V + (\pi r ^2) \frac{dB}{dt}}{B (2 \pi r)}$$

So I need to take the derivative of $B$ with respect to $t$.

$$\frac{dB}{dt} = - 0.2 \sin (4t) (4)$$
$$\frac{dB}{dt} = - 0.8 \sin (4t)$$

Then

$$v = \frac{V - 0.8 \sin(4t) (\pi r^2)}{B (2 \pi r)}$$

Replacing with the values of the statement (r=0.5m)
$$v = \frac{V - 0.8 \sin(4t) (\pi 0.5^2)}{(0.2 \cos(4t)) (2 \pi 0.5)}$$
$$v = \frac{V - 0.2 \pi \sin(4t)}{0.2 \pi \cos (4t)}$$

From the statement I know that when radius is 0.5m, the RMS induced voltage is 0.5V, so I can do:

$$V_{\text{RMS}} = \frac{V_{\text{peak}}}{\sqrt{2}}$$
$$V_{\text{peak}} = \sqrt{2} V_{\text{RMS}}$$

Can I just replace in $V$?
$$v = \frac{ 0.5 \sqrt{2} - 0.2 \pi \sin(4t)}{0.2 \pi \cos (4t)}$$

I need the parameter $t$ to correspond for when $V$ takes its peak value. But I just do not see the relation... I do not see how can I get rid of the $t$.

I have been playing with the graphs of the situation and I found that if I want $V_{\text{peak}}$ to be equals to $\sqrt{2} V_{\text{RMS}} = 0.5 \sqrt{2}$ I need the velocity $v$ to be approximately equals to 0.51625 and in consecuence $t=0.2736$

The green curve is the curve that correspond to the voltage $V = vB (2 \pi r) - \frac{dB}{dt} (\pi r ^2)$. Desmos Link.

View attachment 244375

But without the graph and the approximations I am lost with the equation

$$v = \frac{ 0.5 \sqrt{2} - 0.2 \pi \sin(4t)}{0.2 \pi \cos (4t)}$$

I tought that maybe I could find $t$ by finding what value of $t$ verifies $\frac{dV}{dt} = 0$. But for that I would need the value of velocity $v$...

Okay, I see what you did now with your steps,

$v = \frac{V - 0.8 \sin(4t) (\pi 0.5^2)}{(0.2 \cos(4t)) (2 \pi 0.5)}$

$v = \frac{V - 0.2 \pi \sin(4t)}{0.2 \pi \cos (4t)}$

That looks correct to me now. So far so good.

But then when you find the RMS values, I don't think you are doing that correctly.

Let me give you an example. If

$V(t) = V_{peak} \cos (\omega t),$

then

$V_{rms} = \frac{V_{peak}}{\sqrt{2}}.$

Note that the sinusoid goes away along with the t.

I mean that $V_{rms} \neq \frac{V_{peak}}{\sqrt{2}} \cos (\omega t),$ rather it's just $\frac{V_{peak}}{\sqrt{2}}$.

As soon you divide things by $\sqrt{2}$ to switch over to RMS, there shouldn't be any sines or cosines left.

 

[adrms]

Okay, I see what you did now with your steps,

$v = \frac{V - 0.8 \sin(4t) (\pi 0.5^2)}{(0.2 \cos(4t)) (2 \pi 0.5)}$

$v = \frac{V - 0.2 \pi \sin(4t)}{0.2 \pi \cos (4t)}$

That looks correct to me now. So far so good.

But then when you find the RMS values, I don't think you are doing that correctly.

Let me give you an example. If

$V(t) = V_{peak} \cos (\omega t),$

then

$V_{rms} = \frac{V_{peak}}{\sqrt{2}}.$

Note that the sinusoid goes away along with the t.

I mean that $V_{rms} \neq \frac{V_{peak}}{\sqrt{2}} \cos (\omega t),$ rather it's just $\frac{V_{peak}}{\sqrt{2}}$.

As soon you divide things by $\sqrt{2}$ to switch over to RMS, there shouldn't be any sines or cosines left.

Oh, I see...

So the equation of the voltage signal is:
$$V = vB (2 \pi r) - \frac{dB}{dt} (\pi r ^2)$$
$$V = v(0.2 \cos (4t)) (2 \pi r) + 0.8 \sin (4t) (\pi r ^2)$$

Replacing the radius r=0.5m

$$V = 0.2 \pi v \cos (4t) + 0.2 \pi \sin (4t)$$

Then I can "merge" the sinusoidal terms using trigonometric properties
$$V = 0.2 \pi (v \cos (4t) + \sin (4t))$$
$$V = 0.2 \pi (\sqrt{v^2 + 1} \sin (\alpha) \cos (4t) + \sqrt{v^2+1} \cos (\alpha) \sin (4t))$$
$$V = 0.2 \pi \sqrt{v^2 + 1} (\sin (\alpha) \cos (4t) + \cos (\alpha) \sin (4t))$$
$$V = 0.2 \pi \sqrt{v^2 + 1} (\sin (4t + \alpha))$$

So now I can see that

$$V_{\text{peak}} = 0.2 \pi \sqrt{v^2 + 1}$$

Now I can use the information abour $V_{\text{RMS}}$
$$V_{\text{RMS}} = \frac{V_{\text{peak}}}{\sqrt{2}}$$
$$V_{\text{peak}} = \sqrt{2} V_{\text{RMS}}$$
$$0.2 \pi \sqrt{v^2 + 1}=0.5 \sqrt{2}$$
$$\sqrt{v^2 +1} = \frac{0.5 \sqrt{2}}{0.2 \pi}$$
$$v^2 +1=\left( \frac{0.5 \sqrt{2}}{0.2 \pi} \right) ^2$$
$$v =\sqrt{\left( \frac{0.5 \sqrt{2}}{0.2 \pi} \right) ^2 - 1}$$
$$v =0.51625$$

And that is the value that I have found previously with the graphic, so I suppose it is correct. Right?

I really appreciate your help @collinsmark Thank you very much for your guidance.
@scottdave your question was really helpful to make me rethink the relation between the signal and its RMS value. Thank you very much.

 

[collinsmark]

[Edit: on further analysis, the OP's answer looks correct to me after the OP outlined the steps in more detail. See the next two posts.]

@adrms,

Honestly, you lost me on that last post. I still think there is a misunderstanding about how to determine the rms values of things.

Let me lay out a few more practice examples.

Practice example 1:

$x(t) = \cos(t)$

What is $x_{{\rm{peak}}}$?

What is $x_{{\rm{rms}}}$?

Practice example 2:

$y(t) = \sin( \omega t)$

What is $y_{{\rm{peak}}}$?

What is $y_{{\rm{rms}}}$?

Practice example 3:

$z(t) = A \sin( \omega t)$

What is $z_{{\rm{peak}}}$?

What is $z_{{\rm{rms}}}$?

Practice example 4:

$x(t) = 0.2 \cos( 4 t)$

What is $x_{{\rm{peak}}}$?

What is $x_{{\rm{rms}}}$?

-------
Answers:
practice example 1: $x_{\rm{peak}} = 1, x_{\rm{rms}} = \frac{1}{\sqrt{2}}$
practice example 2: $y_{\rm{peak}} = 1, y_{\rm{rms}} = \frac{1}{\sqrt{2}}$ (same as before)
practice example 3: $z_{\rm{peak}} = A, z_{\rm{rms}} = \frac{A}{\sqrt{2}}$
practice example 3: I'll leave this answer for you to find.

 

[adrms]

@adrms,

Honestly, you lost me on that last post. I still think there is a misunderstanding about how to determine the rms values of things.

Let me lay out a few more practice examples.

Practice example 1:

$x(t) = \cos(t)$

What is $x_{{\rm{peak}}}$?

What is $x_{{\rm{rms}}}$?

Practice example 2:

$y(t) = \sin( \omega t)$

What is $y_{{\rm{peak}}}$?

What is $y_{{\rm{rms}}}$?

Practice example 3:

$z(t) = A \sin( \omega t)$

What is $z_{{\rm{peak}}}$?

What is $z_{{\rm{rms}}}$?

Practice example 4:

$x(t) = 0.2 \cos( 4 t)$

What is $x_{{\rm{peak}}}$?

What is $x_{{\rm{rms}}}$?

-------
Answers:
practice example 1: $x_{\rm{peak}} = 1, x_{\rm{rms}} = \frac{1}{\sqrt{2}}$
practice example 2: $y_{\rm{peak}} = 1, y_{\rm{rms}} = \frac{1}{\sqrt{2}}$ (same as before)
practice example 3: $z_{\rm{peak}} = A, z_{\rm{rms}} = \frac{A}{\sqrt{2}}$
practice example 3: I'll leave this answer for you to find.

I thought that I had understood it well, please let me explain in detail my procedure to solve the problem.

First I will answer the practice example 4 to be sure that I understand where RMS value of a signal comes from.
Let me see, I can derive it seeing the other examples.
$$x_{\text{peak}} = 0.2$$
$$x_{\text{RMS}} = \frac{0.2}{\sqrt{2}}$$

But I know how to analyze it.
By definition I think that RMS value of a signal is
$$x_{\text{RMS}} = \sqrt{ \frac{1}{T} \int _0 ^{T} x(t)^2 dx}$$
The signal of practice example 4 have a period $T = \frac{\pi}{2}$. I know that because the function $\cos (x)$ has a period $T = 2 \pi$ then if you multiply the argument of the function by 4 you are making it 4 times smaller.
$$x_{\text{RMS}} = \sqrt{ \frac{1}{\frac{\pi}{2}} \int _0 ^{\frac{\pi}{2}} (0.2 \cos (4t))^2 dx}$$
$$x_{\text{RMS}} = 0.2 \sqrt{ \frac{1}{\frac{\pi}{2}} \int _0 ^{\frac{\pi}{2}} \cos^2 (4t) dx}$$
$$x_{\text{RMS}} = 0.2 \sqrt{ \frac{1}{\frac{\pi}{2}} \int _0 ^{\frac{\pi}{2}} \frac{1+ \cos (2 \cdot 4x)}{2} dx}$$
$$x_{\text{RMS}} = 0.2 \sqrt{ \frac{2}{\pi} \frac12 \int _0 ^{\frac{\pi}{2}} (1+ \cos (2 \cdot 4x)) dx}$$
$$x_{\text{RMS}} = 0.2 \sqrt{ \frac{1}{\pi} \left( \int _0 ^{\frac{\pi}{2}} 1 dx + \int _0 ^{\frac{\pi}{2}} \cos (2 \cdot 4x) dx \right) }$$
$$x_{\text{RMS}} = 0.2 \sqrt{ \frac{1}{\pi} \left( \frac{\pi}{2} + \frac18 \left( \sin \left( 8 \cdot \frac{\pi}{2} \right) - \sin \left( 8 \cdot 0 \right) \right) \right) }$$
$$x_{\text{RMS}} = 0.2 \sqrt{ \frac{1}{\pi} \frac{\pi}{2} }$$
$$x_{\text{RMS}} = \frac{0.2}{\sqrt{2}}$$

So seeing these examples I thought that there is a pattern... If you have a signal with a sinusoidal form for example.

$$y(t) = A \sin(\omega t + b)$$

The peak and RMS values are:

$$x_{\text{peak}} = A$$
$$x_{\text{RMS}} = \frac{A}{\sqrt{2}}$$

Then in my problem I have a loop that is moving in a uniform magnetic field and it is varying its area. So I know that the induced voltage will be determined by

$$V = \int \left(\vec{v} \times \vec{B}\right) \bullet \vec{dl} - \oint _S \frac{\vec{dB}}{dt} \bullet \vec{ds}$$

From the statement it can be deduced which vectors are parallel with others and which vectors are perpendicular with others

$$V = \int vBdl - \oint _S \frac{dB}{dt} ds$$

The magnetic inductance is varying respect time but it is uniformly applied on the area so at certain moment $t$ it can be considered constant and can be separated from the integrand. The same happens with its derivative.

$$V = vB \int dl - \frac{dB}{dt} \oint _S ds$$
$$V = vB (2 \pi r) - \frac{dB}{dt} (\pi r ^2)$$
$$V = v(0.2 \cos (4t)) (2 \pi r) + 0.8 \sin (4t) (\pi r ^2)$$

Now I replace the value of $r = 0.5$

$$V = 0.2 \pi v \cos (4t) + 0.2 \pi \sin (4t)$$

I have a sinusoidal signal V, and I thought it could be well if I can transform my signal so it could have the form
$$V(t) = A \sin(\omega t + B)$$
Then I do what I did in my last post

$$V = 0.2 \pi v \cos (4t) + 0.2 \pi \sin (4t)$$
$$V = 0.2 \pi (v \cos (4t) + \sin (4t))$$

I say that "there is" a right triangle where one side length is $v$ and the other side length is $1$. The angle between these sides is alpha. And the hypotenuse length is $\sqrt{v^2 + 1}$

Then I can replace V and 1.

$$v=\sqrt{v^2 + 1} \sin (\alpha) $$
$$1 = \sqrt{v^2+1} \cos (\alpha)$$
$$V = 0.2 \pi \left( \sqrt{v^2 + 1} \sin (\alpha) \cos (4t) + \sqrt{v^2+1} \cos (\alpha) \sin (4t) \right)$$
$$V = 0.2 \pi \sqrt{v^2 + 1} (\sin (\alpha) \cos (4t) + \cos (\alpha) \sin (4t))$$

Now I use a trigonometric identity that says $\sin \alpha \cos \beta + \cos \alpha \sin \beta = \sin (\alpha + \beta)$

$$V = 0.2 \pi \sqrt{v^2 + 1} (\sin (4t + \alpha))$$

Now I have the signal V expressed in such way that allows me to see clearly that the peak value must be $0.2 \pi \sqrt{v^2 + 1}$

Knowing that for a sinusoidal wave (including this case) the RMS value is

$$V_{\rm{rms}} = \frac{V_{\rm{peak}}}{\sqrt{2}} $$
$$V_{\rm{peak}} = V_{\rm{rms}} \sqrt{2} $$

The value of RMS voltage is given in the statement. I can replace the variables and find $v$.

$$0.2 \pi \sqrt{v^2 + 1}=0.5 \sqrt{2}$$
$$\sqrt{v^2 +1} = \frac{0.5 \sqrt{2}}{0.2 \pi}$$
$$v^2 +1=\left( \frac{0.5 \sqrt{2}}{0.2 \pi} \right) ^2$$
$$v =\sqrt{\left( \frac{0.5 \sqrt{2}}{0.2 \pi} \right) ^2 - 1}$$
$$v = 0.51625 \left[ \frac{\text{m}}{\text{s}} \right]$$

 

[collinsmark]

I thought that I had understood it well, please let me explain in detail my procedure to solve the problem.

First I will answer the practice example 4 to be sure that I understand where RMS value of a signal comes from.
Let me see, I can derive it seeing the other examples.
$$x_{\text{peak}} = 0.2$$
$$x_{\text{RMS}} = \frac{0.2}{\sqrt{2}}$$

But I know how to analyze it.
By definition I think that RMS value of a signal is
$$x_{\text{RMS}} = \sqrt{ \frac{1}{T} \int _0 ^{T} x(t)^2 dx}$$
The signal of practice example 4 have a period $T = \frac{\pi}{2}$. I know that because the function $\cos (x)$ has a period $T = 2 \pi$ then if you multiply the argument of the function by 4 you are making it 4 times smaller.
$$x_{\text{RMS}} = \sqrt{ \frac{1}{\frac{\pi}{2}} \int _0 ^{\frac{\pi}{2}} (0.2 \cos (4t))^2 dx}$$
$$x_{\text{RMS}} = 0.2 \sqrt{ \frac{1}{\frac{\pi}{2}} \int _0 ^{\frac{\pi}{2}} \cos^2 (4t) dx}$$
$$x_{\text{RMS}} = 0.2 \sqrt{ \frac{1}{\frac{\pi}{2}} \int _0 ^{\frac{\pi}{2}} \frac{1+ \cos (2 \cdot 4x)}{2} dx}$$
$$x_{\text{RMS}} = 0.2 \sqrt{ \frac{2}{\pi} \frac12 \int _0 ^{\frac{\pi}{2}} (1+ \cos (2 \cdot 4x)) dx}$$
$$x_{\text{RMS}} = 0.2 \sqrt{ \frac{1}{\pi} \left( \int _0 ^{\frac{\pi}{2}} 1 dx + \int _0 ^{\frac{\pi}{2}} \cos (2 \cdot 4x) dx \right) }$$
$$x_{\text{RMS}} = 0.2 \sqrt{ \frac{1}{\pi} \left( \frac{\pi}{2} + \frac18 \left( \sin \left( 8 \cdot \frac{\pi}{2} \right) - \sin \left( 8 \cdot 0 \right) \right) \right) }$$
$$x_{\text{RMS}} = 0.2 \sqrt{ \frac{1}{\pi} \frac{\pi}{2} }$$
$$x_{\text{RMS}} = \frac{0.2}{\sqrt{2}}$$

So seeing these examples I thought that there is a pattern... If you have a signal with a sinusoidal form for example.

$$y(t) = A \sin(\omega t + b)$$

The peak and RMS values are:

$$x_{\text{peak}} = A$$
$$x_{\text{RMS}} = \frac{A}{\sqrt{2}}$$

Then in my problem I have a loop that is moving in a uniform magnetic field and it is varying its area. So I know that the induced voltage will be determined by

$$V = \int \left(\vec{v} \times \vec{B}\right) \bullet \vec{dl} - \oint _S \frac{\vec{dB}}{dt} \bullet \vec{ds}$$

From the statement it can be deduced which vectors are parallel with others and which vectors are perpendicular with others

$$V = \int vBdl - \oint _S \frac{dB}{dt} ds$$

The magnetic inductance is varying respect time but it is uniformly applied on the area so at certain moment $t$ it can be considered constant and can be separated from the integrand. The same happens with its derivative.

$$V = vB \int dl - \frac{dB}{dt} \oint _S ds$$
$$V = vB (2 \pi r) - \frac{dB}{dt} (\pi r ^2)$$
$$V = v(0.2 \cos (4t)) (2 \pi r) + 0.8 \sin (4t) (\pi r ^2)$$

Now I replace the value of $r = 0.5$

$$V = 0.2 \pi v \cos (4t) + 0.2 \pi \sin (4t)$$

I have a sinusoidal signal V, and I thought it could be well if I can transform my signal so it could have the form
$$V(t) = A \sin(\omega t + B)$$
Then I do what I did in my last post

$$V = 0.2 \pi v \cos (4t) + 0.2 \pi \sin (4t)$$
$$V = 0.2 \pi (v \cos (4t) + \sin (4t))$$

I say that "there is" a right triangle where one side length is $v$ and the other side length is $1$. The angle between these sides is alpha. And the hypotenuse length is $\sqrt{v^2 + 1}$

View attachment 244396

Then I can replace V and 1.

$$v=\sqrt{v^2 + 1} \sin (\alpha) $$
$$1 = \sqrt{v^2+1} \cos (\alpha)$$
$$V = 0.2 \pi \left( \sqrt{v^2 + 1} \sin (\alpha) \cos (4t) + \sqrt{v^2+1} \cos (\alpha) \sin (4t) \right)$$
$$V = 0.2 \pi \sqrt{v^2 + 1} (\sin (\alpha) \cos (4t) + \cos (\alpha) \sin (4t))$$

Now I use a trigonometric identity that says $\sin \alpha \cos \beta + \cos \alpha \sin \beta = \sin (\alpha + \beta)$

$$V = 0.2 \pi \sqrt{v^2 + 1} (\sin (4t + \alpha))$$

Now I have the signal V expressed in such way that allows me to see clearly that the peak value must be $0.2 \pi \sqrt{v^2 + 1}$

Knowing that for a sinusoidal wave (including this case) the RMS value is

$$V_{\rm{rms}} = \frac{V_{\rm{peak}}}{\sqrt{2}} $$
$$V_{\rm{peak}} = V_{\rm{rms}} \sqrt{2} $$

The value of RMS voltage is given in the statement. I can replace the variables and find $v$.

$$0.2 \pi \sqrt{v^2 + 1}=0.5 \sqrt{2}$$
$$\sqrt{v^2 +1} = \frac{0.5 \sqrt{2}}{0.2 \pi}$$
$$v^2 +1=\left( \frac{0.5 \sqrt{2}}{0.2 \pi} \right) ^2$$
$$v =\sqrt{\left( \frac{0.5 \sqrt{2}}{0.2 \pi} \right) ^2 - 1}$$
$$v = 0.51625 \left[ \frac{\text{m}}{\text{s}} \right]$$

Now I see what you did there. That looks correct* to me. Good job!

*(Well, "correct" in the sense that it may be the answer your coursework is looking for. But this problem is a little ill-conceived due to the use of RMS on a non-stationary signal.)

[ps., forgive me for not seeing this a couple of posts ago. You were correct there too.]