00:01
Part a of the problem, we need to calculate the magnetic field at the location of the coil for t is equal to two seconds.
00:08
The emf induced in the coil is given by e is equal to number of terms, rate of change of magnetic flux, d5b, divided by dt, magnitude, where a phi b, the flux is equal to area times magnetic field.
00:28
Then substituting a b, we can write emf to be a times n, d, b divided by dt.
00:38
From here, then db divided by dt, the rate of change of a tick field is equal to emf divided by a times n.
00:48
The induced emf can be written in terms of induced current and the resistance.
00:52
And we can write db will be divided by dt is equal to r.
00:58
R i divided by a n then db will be equal to so d b is equal to r i divided by a times n d t integrating integrating b will be equal to integral of a d b that is integrating from zero to two seconds r i r i r i a n d d d d d d d d d d d d that is integrating from zero to two seconds r i r i a n d so integrating with respect to i, we get, so this will be r divided by a, n integrating 0 to 2 seconds, i, dt.
01:46
From the given graph we can find the value of the integration by calculating the area under the curve.
01:54
From t is equal to 0 to t is equal to 2 seconds, this will be i, integrate, of 0 to 2 idt is equal to the area that is equal to 1 over 2 2 times 3 milli -ampeer this gives us 0 .003 0mphre seconds then we substitute this value in the equation that we wrote before and find db from 0 to 2 will be 0 .0030 times r divided by an so substituting values for rn which is 0 .250 0 .250 oms times 0 .003 00 divided by by r square is the area which is r is 0 .008 .0 square times four this gives us a db from 0 to 2 to be 0 .935 tesla from 0 to 2 seconds.
03:06
Part b of the problem.
03:08
Now we calculate the magnetic field at location of the coil for t is equal to 5 seconds.
03:15
This will be equal to so db magnetic field from 0 to 5 will be equal to magnetic field from 0 to 2 plus magnetic field from 2 to 5.
03:29
So we'll calculate 2 to 5 db that is 2 to 5 will be equal to r times using the formula that we wrote previously is the area under the curve so area under the curve divided by a times n so we substitute a value 0 .250 ohms is the resistance times the area under the curve for 2 to 5 will be 3 times central power minus 3 a times three seconds divided by pi times r 0 .008 square meters times four...