I can't see the image in this post. Maybe it will become visible in a few minutes...PumpkinCougar95 said:This post has made me wonder even more. Should there be an emf in this case too?
https://drive.google.com/open?id=1tuqI1S7juh8iZhb8t2TaPrVm82CaL5mU
I am saying this because the induced E field is stronger the closer you get to the region with changing magnetic field.
And If it does, Doesn't this kind of violate the fact that the flux through the loop should be changing to create an E field?
PumpkinCougar95 said:loop should be changing to create an E field
PumpkinCougar95 said:And If it does, Doesn't this kind of violate the fact that the flux through the loop should be changing to create an E field?
You can have an ## E ## field at locations on a loop where there is zero changing magnetic flux through that loop. That is ok. In traveling around that loop, you would find ## \oint E \cdot ds =0 ##. This doesn't mean that ## E ## needs to be zero everywhere on the loop for the integral to be zero. (Hopefully this answers your question here). This result follows from ## \nabla \times E=-\frac{\partial{B}}{\partial{t}} ##, integrated over an area, along with Stokes theorem. The result is quite exact. ## \\ ## Additional note: In many cases, computing the ## E=E(r) ## and performing a path integral of this ## E ## over even a simple loop like a circle could be quite difficult. Stokes theorem gives this result immediately. Doing the calculation the long way might take a couple of hours or longer to get this zero result.PumpkinCougar95 said:So what do you think?
Charles Link said:You can have an EE E field at locations on a loop where there is zero changing magnetic flux through that loop. That is ok. In traveling around that loop, you would find ∮E⋅ds=0∮E⋅ds=0 \oint E \cdot ds =0 . This doesn't mean that EE E needs to be zero everywhere on the loop for the integral to be zero.
You have two parts of the loop where ## E ## is perpendicular to ## ds ## giving zero. Then you have two radii, ## \gamma_1 ## and ## \gamma_2 ##. ## E(r)=(A)(\frac{dB}{dt})/(2 \pi r ) ##, with ## A=\pi R^2 ##. For one arc, the path length is ## L_1= \gamma_1 \theta ##. For the other arc, the path length is ## L_2=-\gamma_2 \theta ##, ( with a minus sign). ## E_1=C/\gamma_1 ## and ## E_2=C/\gamma_2 ## for the same constant ## C ##. Thereby, ## E_1 L_1 +E_2 L_2=0 ## This one clearly has ## \oint E \cdot ds=0 ##.PumpkinCougar95 said:But ## \oint E \cdot ds =0 ## is NOT true in this case, Even though flux is zero through the loop at all times:
https://drive.google.com/open?id=18tjoiyAjjD1XAXgmZn_3UKgbtMrFZ6zz
PumpkinCougar95 said:If there is a very very big(infinitely big) region of space where ## \frac {dB} {dt} = constant ## what would be the E field at any point? Obviously ## \nabla x E = constant ## but what after that ?
The direction of the induced electric field is always perpendicular to both the direction of the changing magnetic field and the direction of the motion of the conductor or coil. This is known as the right-hand rule.
The direction of the induced electric field is determined by the relative motion between the conductor or coil and the magnetic field. If the motion is parallel to the magnetic field, no induced electric field is produced.
Yes, the direction of the induced electric field can be reversed by changing the direction of the magnetic field or the direction of the motion of the conductor or coil.
The strength of the induced electric field is affected by the strength of the changing magnetic field, the speed of the motion, and the number of turns in the conductor or coil.
The direction of the induced electric field is important because it determines the direction of the induced current, which can have practical applications such as generating electricity in power plants or inducing currents in transformers.