00:01
For this problem on the topic of magnetism, in a figure we are shown a capacitor which has circular plates, which have a radius of 18 centimeters.
00:09
It is connected to a source with emf -e -m -syne -omega -t, where em is 220 volts, and the angular frequency omega is 130 radiance per second.
00:18
We're given the maximum value of the displacement current id to be 7 .6 microampiers, and we want to find the maximum value of the current i in the circuit, the maximum value.
00:30
Of dfid where fi is electric flux through the region between the plates, the separation between the plates, as well as the maximum value of the magnitude of the magnetic field between the plates at a distance of 11 centimeters from the center.
00:45
Now at any instant in time the displacement current id in the gap between the plates equals the conduction current i in the wires.
00:54
And so the maximum current is equal to the maximum displacement current which is given, and this is 7 .6 microampers.
01:13
Now since the displacement current id is equal to epsilon not dfie dt where fi is the electric flux we have that the maximum value of the rate of change of flux, dfi e dt max is equal to the maximum displacement current divided by epsilon not.
01:48
And so this is 7 .6 times 10 to the minus 6 ampiers divided by the constant epsilon not, which is 8 .85 times 10 to the minus 12 ferrets per meter.
02:06
And so we get d5d to be 8 .59 times 10 to the power 5 volt meter per second.
02:27
Next for part c, if we let the plate area be a and the plate separation be d, then the displacement current id is equal to epsilon knot, d5e, dt, which is epsilon not times d by dt and the electric flux can be written as the area a times the electric field e.
02:54
And so we can write this as, epsilon not times a, which are both constants, times d by dt, and the electric field can be replaced by the potential difference v across the plates divided by the separation of the plates d.
03:10
And so this is epsilon not a over d, d, dv, dt.
03:20
Now, the potential difference across the capacitor is the same in magnitude as the emf of the generator.
03:26
So, v is equal to, epsilon m sine omega t and dv d t is therefore omega times epsilon m cosine if we derive take the derivative with respect to t and so the displacement current i d is therefore epsilon not times a times omega times epsilon m divided by d times the cosine of omega t and the maximum displacement current id max is equal to epsilon not a omega epsilon m over d so letting the cosine omega t term ten to one and so rearranging, we get the separation between plates d to be epsilon not, a, omega, epsilon, m divided by the maximum displacement current...