# Electrical induction due to changing magnetic fields

1. Apr 21, 2006

### Greg825

I'm trying to understand why, when, in a solenoid for example, electrical current is magnetically induced, doesnt the newly created magnetic field due to current "cancel out" the effects of the original changing magnetic field. One person I asked suggested resistance, and this sounds completely plausible, but does that mean current can't be magnetically induced in a superconductor? There may be other factors that are immidiately apparent in the equations we're working with but I don't see them. It seems like it would be quite complex, wouldn't one need some way of finding the rate of change of current due to a changing magnetic field? Is it instaneous (if the dB/dt is constant). note: this I'm assuming area is constant (for flux = area * magnetic field)

I suspect my understanding of these concepts is pretty feeble, and so here I am.

edit: mm maybe this would have been better in the "classical physics" section

Last edited: Apr 21, 2006
2. Apr 22, 2006

### rbj

a changing magnetic field creates an electric field that is at a right angle ("curls" around) and a changing electric field creates a magnetic field. restated in the CGS form:

$$\oint_C \mathbf{E} \cdot d\mathbf{l} = -\frac{1}{c} \ \frac{d}{dt} \int_S \mathbf{B} \cdot d\mathbf{A}$$

$$\oint_C \mathbf{B} \cdot d\mathbf{l} = \frac{1}{c} \ \frac{d}{dt} \int_S \mathbf{E} \cdot d \mathbf{A}$$

now there is nothing that states that the $\mathbf{E}$ field created by the $\mathbf{B}$ field cannot, itself, create another $\mathbf{B}$ field. but, it is only changing E or B fields that create the other. how can we create a changing B field that would result in a changing E field that would result in a changing B field of the same functional form that could conceivably cancel out the first? it would have to be of an exponential (or trig) form of function. and would that secondary changing B field be in the same location as the primary changing B field to cancel it out?

if it's trigonemetric (sine or cosine), you can check the signs and see that, rather than cancel out, the functions team up. someone that is less tired than i am (at 3 a.m.) can figure out how to take the pair of vector equations above and turn them into a wave equation.

3. Apr 23, 2006

### Greg825

thanks for the response. I need to better understand Faraday's law of induction and its relations specifically to current before I can fully understand your response.