Where does this term come from? (pulling a wire loop through a B-field)

snatchingthepi
Messages
145
Reaction score
37
Homework Statement
Pulling a hoop through a uniform B-field
Relevant Equations
emf = loopintegral (f_pull dot dl)
I can't for whatever reason figure out where the sin(theta) term is coming from in the attached picture of page 306 of Griffiths' 4th edition EM text. The paragraph says it comes from the dot product, but I just don't see where it's coming from.
2019-09-10.png
 
Physics news on Phys.org
Can you also scan the figure that this is referring to? Theta must be the angle of the loop with respect to the B-field direction?
 
berkeman said:
Can you also scan the figure that this is referring to? Theta must be the angle of the loop with respect to the B-field direction?

Yes here it is.
 

Attachments

  • 2019-09-10 (1).png
    2019-09-10 (1).png
    102.3 KB · Views: 245
  • 2019-09-10 (2).png
    2019-09-10 (2).png
    121.4 KB · Views: 240
So the dot product enters in because if the coil is oriented parallel to the B-field, then none of the flux pierces the plane of the loop. Does that make sense?
 
  • Like
Likes snatchingthepi
yes it all fits now, ##\theta## is the angle between ##\vec{u}## (the drift velocity of a charge inside the conductor) and the total velocity of the charge ##\vec{w}##. The element ##\vec{dl}## of the integral is in the direction of the total velocity ##\vec{w}## (it is ##\vec{dl}=\vec{w}dt##), because as Griffith says to find the work of ##F_p## we have to follow a charge at its journey around the loop, and this journey is done with the total velocity ##\vec{w}##. Thus the angle between ##\vec{dl}## and ##\vec{F_p}## is the angle between ##\vec{w}## and ##\vec{F_p}## which is ##\frac{\pi}{2}-\theta## .Thus and by definition of dot product ##\vec{F_p}\cdot \vec{dl}=F_pdl\cos(\vec{F_p},\vec{dl})=F_pdl\cos(\frac{\pi}{2}-\theta)=F_pdl\sin\theta##
 
Last edited:
  • Like
Likes snatchingthepi
Thank you all!
 
  • Like
Likes berkeman and Delta2
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top