A circular parallel plate capacitor with plate radius 8.2cm and plate separation .15mm is connected in series with a 16V ideal battery and a 790 ohm resistor at t=0. At time t=2.4e-6 seconds, what is the induced magnetic field that lies along the outer edge of the capacitor?

I found current, but how do I even get the magnetic field??

A circular parallel plate capacitor with plate radius 8.2cm and plate separation .15mm is connected in series with a 16V ideal battery and a 790 ohm resistor at t=0. At time t=2.4e-6 seconds, what is the induced magnetic field that lies along the outer edge of the capacitor?

I found current, but how do I even get the magnetic field??

To find the induced magnetic field that lies along the outer edge of the capacitor, you can use the Biot-Savart law. The Biot-Savart law relates the magnetic field created by a current-carrying conductor to the magnitude and direction of the current.

Now, to use the Biot-Savart law, we need to determine the current flowing through the circular parallel plate capacitor. Since the capacitor is connected in series with a resistor and an ideal battery, the current can be found using Ohm's Law, which states that V = IR, where V is the voltage, I is the current, and R is the resistance.

Here's how to find the current flowing through the circuit:

1. Calculate the total resistance (RT) of the circuit by summing up the resistance of the resistor (R) and the equivalent resistance of the capacitor (RC). The equivalent resistance of the capacitor can be calculated using the formula RC = (2πε₀r) / d, where ε₀ is the permittivity of free space, r is the radius of the capacitor plates, and d is the plate separation.

RC = (2π * 8.85x10^-12 F/m * (0.082m) ) / (0.00015m)
RC ≈ 0.00787 Ω

RT = R + RC
= 790 Ω + 0.00787 Ω
= 790.00787 Ω (approximately)

2. Substitute the given voltage (V = 16 V) and the total resistance (RT) into Ohm's Law (V = IR) to find the current (I).

I = V / RT
= 16 V / 790.00787 Ω
= 0.02027 A (approximately)

Now that we have the current flowing through the circuit, we can proceed to calculate the induced magnetic field along the outer edge of the capacitor using the Biot-Savart law.

The Biot-Savart law states that the magnetic field (dB) at a point due to a small segment of current-carrying wire (dl) is given by the formula: dB = (μ₀ / 4π)(Idl × r̂) / r², where μ₀ is the permeability of free space, Idl is the current element, r̂ is the unit vector pointing from the current element to the point of interest, and r is the distance between the current element and the point of interest.

To find the magnetic field at the outer edge of the capacitor, we need to consider the contribution from all the current elements around the edge. We can assume a circular path along the outer edge of the capacitor and calculate the magnetic field due to each small segment of the current-carrying wire.

However, since the plates of the capacitor are parallel, the electric field within them will be zero. Hence, there is no current flowing along the surface of the plates, resulting in no magnetic field on the outer edge of the capacitor.

Therefore, the induced magnetic field along the outer edge of the capacitor is zero.