00:01
Okay, so we know magnetic flux can be equal to integral bda.
00:04
B is magnetic field, a is area, and b can be expressed as mu -0 -0 -i over 2xr.
00:09
A can be equal to l -dr.
00:11
So if we use mu -0 -0 -2 -r and l -d -r substitute for bda, we'll have 5b is equal to the integral integral fringe from h to h -2 -w, and what's inside as mu -0 -0 -2 -r times l -t, which is also equal to mu -0 -1 -2 -5 times the integral.
00:29
And we know the anti -durative for 1 over r is natural log r.
00:35
Therefore, if we do some arrangement here, eventually you have the 5b is equal to mu 0 -il over 2 pi, 10 natural log h plus w over h.
00:45
So now let's take look at part b.
00:47
So we know the current equation is given as a plus b.
00:50
Therefore, d -i over d -t should be equal to b...