Talk:QB/d cp2.13

Magnetic flux ${\displaystyle \Phi _{m}=\int _{S}{\vec {B}}\cdot {\hat {n}}dA}$
Motional ${\displaystyle \varepsilon =B\ell v}$ if ${\displaystyle {\vec {v}}\perp {\vec {B}}}$
Electromotive "force" (volts) ${\displaystyle \varepsilon =-N{\tfrac {d\Phi _{m}}{dt}}=\oint {\vec {E}}\cdot d{\vec {\ell }}}$
rotating coil ${\displaystyle \varepsilon =NBA\omega \sin \omega t}$

Question 7

A spatially uniform magnetic points in the z-direction and oscillates with time as ${\displaystyle B(t)={\vec {B}}_{0}\sin \omega t}$ where ${\displaystyle B_{0}=1.5T}$ and ${\displaystyle \omega =2.0E3s^{-1}}$. Suppose the electric field is always zero at point O, and consider a circle of radius 0.5 m that is centered at that point and oriented in a plane perpendicular to the magnetic field. Evaluate the maximum value of the line integral ${\displaystyle \oint {\vec {E}}\cdot d{\vec {\ell }}}$ around the circle.

Note: There are three common variables of integration for these line integrals: ${\displaystyle \int {\vec {F}}\cdot d{\vec {\ell }}}$, ${\displaystyle \int {\vec {F}}\cdot d{\vec {s}}}$, and ${\displaystyle \int {\vec {F}}\cdot d{\vec {r}}}$ are all equivalent (provided the paths are the same).

Solution

Since the magnetic field is parallel to the area of the circle, the magnetic flux is, ${\displaystyle \Phi _{m}=\pi R^{2}B_{0}\sin \omega t}$. Although we have introduced Faraday's law in terms of the emf,

${\displaystyle \epsilon =-{\frac {d}{dt}}\Phi _{m}\,,}$

recall that voltage and electric field have been related by,

${\displaystyle V_{B}-V_{A}=-\int _{A}^{B}{\vec {E}}\cdot d{\vec {\ell }}.}$

For reasons beyond the scope of this course, the electromotive force is defined without this minus sign, so that:

${\displaystyle \oint {\vec {E}}\cdot d{\vec {\ell }}=\epsilon ={\frac {d}{dt}}\Phi _{m}={\frac {d}{dt}}\pi R^{2}B_{0}\sin \omega t}$

We need not be concerned about the minus sign because we seek the maximum value, which is:

${\displaystyle \pi R^{2}B_{0}\omega \cos(0)=\omega \pi R^{2}B_{0}}$ (=2E3*pi*.5^2*1.5=2356 volts)

The stated answer 9.425E+03 V is off by a factor of 4.

Question 8

The induced electric field caused by a solenoid is calculated differently, depending on whether one is inside or outside the solenoid.

A long solenoid has a radius of 0.7 m and 50 turns per meter; its current decreases with time according to ${\displaystyle I_{0}e^{-\alpha t}}$where ${\displaystyle I_{0}=3A}$ and ${\displaystyle \alpha =.25s^{-1}}$. What is the induced field at a distance 2.0 meters from the axis at time t= 0.04s?

Solution

First we find dB/dt where B is the magnetic field:

${\displaystyle B=\mu _{0}nI_{0}e^{-\alpha t}\Rightarrow {\frac {dB}{dt}}=-\alpha \mu _{0}nI_{0}e^{-\alpha t}}$

where ${\displaystyle \mu _{0}=4\pi \times 10^{-7}Tm/A}$ and ${\displaystyle n}$ is the number of turns per meter in the long solenoid.

Since 0.7 meters is less than 2.0 meters, we calculate the magnetic flux using the cross sectional area of the solenoid, and not that associated with the line integral. It is helpful to define two radii: ${\displaystyle R_{s}=R_{solenoid}=.7m}$ and ${\displaystyle r_{f}=r_{field}=2m}$, where the latter describes the distance to the point where the (electric) field is to be calculated. The magnetic flux obeys:

${\displaystyle |\epsilon |=\left|{\frac {d}{dt}}\Phi \right|=\left|\pi R_{s}^{2}{\frac {dB}{dt}}\right|=\pi R_{s}^{2}\cdot \alpha \mu _{0}I_{0}e^{-\alpha t}}$

Calculating the line integral we have:

${\displaystyle |\epsilon |=\oint {\vec {E}}\cdot d{\vec {\ell }}=2\pi r_{f}E}$

Hence we have an electric field that falls as the inverse of the distance to the axis:

${\displaystyle E={\frac {\pi R_{s}^{2}\cdot \alpha \mu _{0}nI_{0}e^{-\alpha t}}{2\pi r_{f}}}={\frac {(0.7^{2})(25)(1.257E-6)(50)(3)(e^{-(25)(.04)})}{(2)(2)}}}$

A. 2.124E-04 V/m

Question 9

9. A long solenoid has a radius of 0.7 m and 50 turns per meter; its current decreases with time according to I0e −αt, where I0 =3 A and α =25 s−1 .What is the induced electric fied at a distance 0.15 m from the axis at time t=0.04 s ?9

Solution

A. 1.300E-04 V/m B. 1.430E-04 V/m C. 1.573E-04 V/m D. 1.731E-04 V/m E. 1.

Example 13.9 is at best confusing

Example 13.9 purports to calculate the average emf over a quarter-cycle of a single-pole generator. The formula they produce is:

${\displaystyle {\langle emf\rangle }_{\text{OpenStax}}=NAB{\frac {\pi /2}{\tau }}={\frac {\pi /2}{\tau }}\Phi _{0}}$

where ${\displaystyle \Phi _{0}=NAB}$ is the peak flux and ${\displaystyle \tau =T/4}$ is a quarter-period. But period and angular frequency are related by,

${\displaystyle \omega T=2\pi \Rightarrow \omega \tau =\pi /2}$, and hence:
${\displaystyle {\langle emf\rangle }_{\text{OpenStax}}=\omega \Phi _{0}=\omega NAB=\epsilon _{0}}$,

which is the peak emf. The "average value" is a vague term, but here is how I would calculate the time average over a quarter cycle (which is not the square root of the rms value):

${\displaystyle \epsilon (t)=\Phi _{0}\sin(\omega t)}$
${\displaystyle \langle emf\rangle _{t}={\frac {\Phi _{0}}{\tau }}\int _{0}^{\tau }\sin(\omega t)dt={\frac {\Phi _{0}}{\omega \tau }}\int _{0}^{t=\tau }\sin(\omega t)d(\omega t)}$

Now change the variable of integration to ${\displaystyle \theta =\omega t}$ and use ${\displaystyle \omega \tau =\pi /2}$:

${\displaystyle \langle emf\rangle _{t}={\frac {\Phi _{0}}{\tau }}\int _{0}^{\omega \tau }\sin(\theta )d\theta ={\frac {2}{\pi }}\omega \Phi _{0}\int _{0}^{\pi /2}\sin(\theta )d\theta ={\frac {2}{\pi }}\epsilon _{0}}$

Since "average" is a vague term, I suppose ${\displaystyle {\langle emf\rangle }_{\text{OpenStax}}}$ can be called some sort of average, but it seems artificial to me.

Why I simplified Example 13.5

It is important to avoid difficult problems on tests and quizzes if there is a formula that students can easily remember -- unless we construct the course so that they are not tempted to memorize the formula

I substituted this one for the rotating rod problem of OpenStax University Physics II Example 13.5 because while the example makes for a good physics lesson, it's not about the equation, but how the equation was derived. Example 13.5 will make a good but very challenging exam question only when we have dozens of similar questions all modified in such a way that equation memorization is rendered impossible for most students (I'm sure there will always be students with photographic memories, but they have special talents that require individual attention -- attention we could afford to give them if we can reduce the cost to educate your typical autodidactic calculus-based physics student.)

A cylinder of height 1.1 cm and radius 3.1 cm is cut into a wedge as shown. Now imagine that the volume grows as θ increases while the radius R and height h remains constant. What is the volume's rate of change if point P is 2.1 cm from point O and moves at a speed of 5.1 cm/s? Assume that the wedge grows in such a way as the front face moves by rotating around the axis (that contains point O.)

-a) 8.767E+00 cm3/s
-b) 9.644E+00 cm3/s
-c) 1.061E+01 cm3/s
-d) 1.167E+01 cm3/s
+e) 1.284E+01 cm3/s

To solve it, we begin by using proportional reasoning to get the volume of the wedge, which is some fraction of the volume of the cylinder. The fraction in what follows represents what fraction the wedge is of the entire cylinder:

${\displaystyle V=\pi R^{2}h\cdot {\frac {\theta }{2\pi }}={\tfrac {1}{2}}hR^{2}\theta }$

${\displaystyle {\frac {\mathrm {d} V}{\mathrm {d} t}}={\tfrac {1}{2}}hR^{2}\omega ={\tfrac {1}{2}}h{\frac {R^{2}}{r}}v}$ = ${\displaystyle {\frac {(1.1{\text{cm}})(3.1{\text{cm}})^{2}(5.1{\text{cm/s}})}{(2)(2.1{\text{cm}})}}}$ = ${\displaystyle 1.2836\approx 1.28{\text{cm/s}}}$

One of the drawbacks to quizbank is that I have not rounded answers to reasonable values. This was deliberate because it permits students to explore their own rounding practices and policies. Instead of showing students how to round in each problem, let's just have a wikiquiz or two on the subject!--Guy vandegrift (discusscontribs) 21:42, 21 June 2018 (UTC)