00:01
For this problem on the topic of electromagnetic induction, we are told that the magnetic field within a long straight solenoid with a radius r is increasing at the rate of dbt.
00:13
We want to then calculate various properties of this magnetic field.
00:19
We firstly want to calculate in part a the rate of change of flux through a circle with radius r1 within the solenoid.
00:31
Now we know that the rate of change of magnetic flux, d phi b, d t, is equal to the area, the cross -sectional area a, times the rate of change of the magnetic field, d b, d -t.
00:50
Since phi, since it is simply b times a, with the cross -sectional area a being constant.
00:55
Now we know this is a circular cross -sectional area, so this is pi r1 squared where r1 is a radius.
01:03
And this is multiplied by db d t.
01:08
So here we have an expression for the rate of change of magnetic flux through a circular cross -section of the solenoid.
01:21
Now in part b, we want to calculate the magnitude of the induced electric field inside the solenoid at the same distance r1 from the axis.
01:30
We then want to show the direction of this field in a diagram.
01:33
So firstly, to calculate the magnitude of this field, we know the electric field e is equal to 1 over 2 pi r or 2 pi r 1 in this case times the rate of change of flux, d5dt, magnetic flux.
01:53
And so we can write this as pi r1 squared over 2 pi r1.
02:05
Our expression for d5dt comes from our first answer.
02:11
And this is multiplied by d b d t the rate of change of magnetic field.
02:18
And so this becomes r1 over 2 d b d t.
02:27
And that's the magnitude of the electric field.
02:31
And now we need to show the direction...