But if more current flowing then it means more voltage is built up then why can't we double the ratio between the current and the resistance so it will be I^4 R instead?
The power loss is given by I ^2 R where I is the current and R is the resitance.
I know why these variables are in this equation.
V = I R
P = I V
therefore,
P = I² R
What I don't understand is the practical explanations for this. Why are there two current variables? Are there two...
I know that the area will be bigger therefore the flux will be larger, but this doesn't mean that the current will be larger does it? Since the number of loops will be less if the diameter is bigger right? So my conclusion is that the current will be the same since the induced emf is the same...
In an induction experiment a tightly wound solenoid is used to produce a current of 2A. The
solenoid is then unwound to make another tightly wound solenoid of twice the diameter.
What current would you expect to get if you use the new solenoid in the same experiment?
I'm new to this forum...
A rod (shown in the picture on the link) rotates around the point O in a magnetic field. It is said for sure that the induced EMF in the rod is larger at point R compared to point P. Why's that?
The picture: http://img58.imageshack.us/my.php?image=physicskx4.png