GREPhysics.NET: GR9677 #46

GR9677 #46

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

Electromagnetism $\Rightarrow$ }Faraday Law

Recall Faraday Law, $\epsilon = -\frac{d\Phi}{dt}$ , where $\Phi=B\cdot dA$ . Since the magnetic field is constant, the equation simplifies to $\epsilon = -B \cdot \frac{dA}{dt}$ for this case.

$B \cdot A = B \cos(\omega t) \pi r^2$ , and thus $d\Phi/dt = -\omega \sin(\omega t) \pi r^2 = -\epsilon = -\epsilon_0 \sin(\omega t)$ . Solving for angular momentum, one has $\omega = \epsilon_0/(B\pi R^2)$ .

Alternatively, one has $A=\pi r r(t) \Rightarrow \dot{A} = \pi r \dot{r} = \pi r v$ . Since $v=\omega r\sin(\omega t)$ , one has $A=\pi r \omega r\sin(\omega t)$ . Plug it into Faraday Law and solve for angular velocity.

Alternate Solutions

There are no Alternate Solutions for this problem. Be the first to post one!

Comments

dipanshugupta
2017-03-29 10:45:02

How do you get the $B cos(\\omega t)$ term for $B.dA$ ?

Chungha1627
2017-08-25 01:48:45

I think that dot product between B and A makes the cosine term.

Reply to this comment

Poop Loops
2008-11-07 20:35:25

This is such a load of crap. In the question they say the loop rotates with angular speed $\omega$ and then they ask you for the angular speed of the loop. What the hell does that mean???

AER
2009-03-31 15:25:31

They want an expression for $\omega$ . So the answer is: $\omega=\epsilon_0 / (B \pi R^2)$

Reply to this comment

bpat
2007-10-30 19:25:37

I think dimensional analysis works fastest.

w = [rads/s]
emf = B*Area/time

so: w = emf/(B*Area)

Only one choice that looks like that and it is (C).

bpat
2007-10-30 19:28:01

w = [rads/s] which is the same thing as [1/s]

Camoph
2010-03-30 18:50:20

Is faster..thanks

flyboy621
2010-10-24 16:14:51

nicely done!

RusFortunat
2015-10-20 17:21:52

брат братан братишка

Reply to this comment

Richard
2007-09-23 23:07:51

$V_{0}{\sin(\omega{t})}=V$
Integrate this to get the flux as a function of time, with an extra negative one (Faraday's law):

$\Phi_{b}(t)=\frac{V_{0}}{\omega}{\cos(\omega{t})}$

At $t=0$ you have the maximum flux which is $\pi{R^2}B$ by the diagram.

Thus, $\omega=V_0/{\pi{R^2}B$

Reply to this comment

Richard
2007-09-23 22:53:49

I don't think you meant to write "angular momentum."

Reply to this comment

Post A Comment!

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.

Bare Basic LaTeX Rosetta Stone