Acapacitor with circular plates of radius 0.0534 m separated by a distance of 4.80 ✕ 10−4 m is being charged by a battery with an emf of 21.0 v. a vacuum exists between the plates of the capacitor. the constructed circuit has a resistance of 450 ω in series with the capacitor, and the capacitor is initially uncharged when a switch is closed, completing the circuit and causing a current to flow (an rc series circuit). find the magnitude of the magnetic field between the plates at a distance of 0.0250 m from the axis through the center of the plates, when one time constant has passed since the switch was closed.

B = 1.3734*10^-7 T

Explanation:

I'm going to take a shot at this since I doubt anyone else would, but I might be wrong, so keep that in mind. Okay, with that out of the way, we know the equation for capacitance:

We also know that it is put in series with a resistor. Since it is initially uncharged and is in series with a battery, it is step response. Using KVL, we will get:

We also know from Ampere's law, there is a magnetic field because of the current induced (right hand rule). From this we can calculate the magnetic field using a contour integral. We can simplify and avoid using Biot-Savart law by using the path of integration of a circle, thus:

Okay now we got all that out of the way, lets first find the area of the plates so we can use the first equation to get capacitance.

Now we need to calculate the current through the capacitor. To do this, use the step response to calculate the voltage drop across the capacitor and then use KVL calculate the current through the resistor. So:

Finally Ampere's law:

B = 1.3734*10^-7 T

pecialized keyboards

answer: d. objects on jupiter experience more gravity due to more gases