A circular circuit loop of radius H consists of a capacitor...

A circular circuit loop of radius H consists of a capacitor of capacitance C, and two inductors with inductance $L_1$ and $L_2$. At $t = 0$ the charge on the top plate of the capacitor is $-Q_0$ while the charge on the bottom plate is $+Q_0$, while the current is $i_0$ flowing in the counterclockwise direction as shown. There is an external magnetic field of magnitude $\beta t$ which points out of the page. a) Starting with some law, find the equation you could solve to find the charge on the capacitor as a function of time. b) Find the charge on the capacitor as a function of time.

Transcript

00:01 Hi there, so for this problem, what we are given, and what we are told is that there is a circular shape circuit of a radius lower case r, containing a resistance capital r in a capacity c.

00:16 Now we are given the voltage difference and increases with some expression in here that between the terminals b and a, it is increasing by the initial value, be sub 0 and these times 1 minus the exponential of minus the time divided by the times tau and how our positive constants.

00:43 So what we need to determine is the rate of change of the magnetic field with respect to time.

00:53 And the last question is is the magnetic field become larger or smaller as time increases? so we we know that in this case, the nph around the loop is equal to the time derivative of the loops, as we know from the equation 29 .2a from the book.

01:14 Now, since the area of the coil is constant, the time derivative of the flusps is equal to the derivative of the magnetic field, multiplied by the area of the loop.

01:23 Now, to calculate the nth in the loop, we add the voltage drop across the capacitor to the voltage drop across the resistor.

01:31 Now the current in the loop is the derivative of the charge on the capacitor.

01:37 So let's start with that.

01:39 We know that the current is the rate of change of the charge with respect to time.

01:46 So we can write this because we know that the charge is the product between the capacitance and the potential.

01:54 Now we know that the capacitance is a constant so we can take it out of the derivative.

01:59 Then we just need to derivate the expression that we are given for the voltage, which is just b -0 times 1 minus the exponential of the time divided by tau.

02:17 If we derive this, the solution for this is then just simply the capacitance times the voltage b -0, this divided by the times tau, this times the exponential of minus the time divided by tau.

02:36 Simplifying this, we will find that this is equal to just simply b -0 divided by the resistance, and this times the exponential of minus the time divided by tau.

02:52 Because remember that tau is defined as just simply the resistance times the capacitance...

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