00:01
Part a of the given problem, the induced electric field is in the direction of induced current, the direction of the magnetic field is into the page and decreasing, so the magnetic flux is also decreasing.
00:18
According to the lens law, the direction of the induced magnetic field must be the same direction of the original magnetic field, which is into the page.
00:27
So according to right handle, the direction of the induced current is clockwise.
00:35
Hence, the direction of induced electric field is a clockwise.
00:42
The electric field is tangent to the ring and its direction is clockwise.
00:47
As we can see from the figure here, let me draw the figure.
00:51
So here we have magnetic field and magnetic field here.
01:00
So from a circle here, so let's say this is the bigger circle with radius r here and the electric field is here inside e with making a radius of a small r here.
01:24
Then for the part a, so the electric field here in the for part a, sorry, is tangent to the ring and its direction is clockwise.
01:38
So here is electric field clockwise.
01:41
Part b, now we need to find the magnitude of the induced electrophil at the r is equal to capital r.
01:48
The induced electric fuel is given by ferrarian's law that says in the ucmph is integral of e .d .l.
01:58
That is a negative rate of a flux change d t.
02:02
The magnetic flux through circle radius r is flux is equal to area times v which is pi r square times b.
02:12
The integration on the left hand side gives us integral of e .d .l is equal to electric field times the part which is a 2 pi r circumference.
02:27
Then we can write electric field e2 pi r is equal to pi r square times the db divided by d t.
02:37
From here, electric field will be r divided by 2, db divided by d t...