Yeah, for this problem on the topic off electromagnetic induction were given a circuit as shown in the figure, and we are also given the capacitance and devoted to which it is initially charged. We have a resistor r no to the given resistance, and we know at time T is equal to zero. The switch is closed. The small circuit diagram is not connected to the large one at all. And we know for the small circuit the resistance per meter on the number of loops for the circuit now the lots ago.
The large circuit is a rectangle on the small. One has given dimensions. Now, if both circles are held stationary and only the wire nearest the small circuit produces an appreciable magnetic field, were asked to calculate the current in the large second Milad circuit 200 microseconds. After the switch is closed, they cut it to be small six circuit at the same time, the direction of the current in the small circuit. And then we were asked to justify why we can ignore the magnetic field from all the wires off the large circuit, except for the closest one now the current in a discharging RC circuit.
I is equal to Vienna over R E to the minus t over RC with the noticed the initial voltage across the capacity now for the small loop. The resistance is the number of loops 25 times 0.6 m times one home per meter, which gives us 15 homes. So let's begin to solve the problem. So in part A, we know the large circuit is an RC circuit and RC circuit has a time constant tall, which is equal to our time. See And this is equal to a resistance which is given as 10 homes time liquor bastions off 20 micro faris, or 20 times 10 to the minus six parents, which gives us a time constant off 200 microseconds.
So the function the current is as a function of time. All right is equal to 100 volts invited by 10 homes to the times e to the minus t over 200 microseconds t over tall. So if we let t b 200 microseconds, we get the current I twenties. They put 200 microseconds to be Yeah, 10 NPR's and E to the minus one, and so this current his three 0.7 and peers. So that's the current in the large circuit.
Any time off 200 microseconds. Now we want to calculate the same for the small circuit. So if we assume that only the long wire nearest the loop produces a magnetic flux through the small loop and we refer to our hint, we get the magnetic flux through the small finally to be the integral from sea. Two c plus a off new Not I be over two pi r D r and all of these are constants, so they come out of the integral. And if we solve this, we get this to be new.
Not I kind of be over two pi times the integral off one of our ER, which becomes the natural log off one that's a of the sea. So therefore, the E M f induced me small look. At T is equal to 200 microseconds. He's epsilon, which is minus in times deep. Rate of change of floods defy B T t.
And we can write this as minus and times you not times be over. Two pi on the natural log off one plus a oversea times. The only thing that changes with time here is the current. So the i d t. Now we can substitute our values into this and we get this to B minus and in 25 tons times may not we know is full pie times 10 to the minus seven in S I units, which is whether for NPR meter squared we'll suppress the units here Times B which were given to be zero point 2 m divided by two pi times the natural long off one glass oversee which is the natural love off three multiplied by minus 3.7 NPR's any time off 200 microseconds or 200 times 10 to the minus six in seconds.
Since all our values I s i units, we get the e m f epsilon being used e m f in s I units to be positive 20 mill evils. So if we have the MF, we confined induced current in the small room. So the induced current in the small will call my prime. I prime is equal to Absalon off our and we've already found the resistance off the small look from above. So this is 20 mil evils off the resistance off 15 homes, which gives us the current inducing the small look to be one 0.33 million and peers.
So now we have the current in the large loop, as well as the current induced in the small loop. Now we want to find the direction off the current in the small circuit. Going to now. The magnetic field from the large loop is directed out of the page within the small loop, so the induced current in the smaller loop will act to oppose the decrease influx from the large loop. And so the induced current flows counterclockwise and finally for party.
We want to know why we can ignore them in eight field from all of the otherwise off the large circuit. Besides the closest one. And the answer to this is that these three wires Milad Luke are too far away to make a significant contribution to the flux in the small look. And we can see this by comparing the distance see to the dimensions off the large loop.
