| DFB |
 |
| Dt |
| D(BA cosq) |
|
| Dt |
A loop of wire has an area of 0.5 m2 and a resistance of R = 0.1 W. There is a uniform magnetic field of B = 1.0 T passing through the loop into the page.
What is the direction of the induced current?
Nothing is changing so there is no induced emf and no induced current.
Now the magnetic field is reduced steadily from 1.0 T to 0 over a 10 second period.
In other words DB/Dt = -0.1 T/s for 10 seconds.
Plot the magnetic flux as a function of time. Plot the induced current as a function of time.
What current is induced in the wire while the field is being reduced?
To find current first find the induced emf:
| e | = | – N |
|
= | – N |
|
In this situation N = 1, and A and q are constants. This gives:
| e | = | – A cos(q) |
|
= | -0.5 x 1 x -0.1 | = | 0.05 V |
The sign doesn't really matter here.
While the magnetic field is changing the loop acts as if it has a battery in it. Using Ohm's Law, the current is:
| I | = |
|
= |
|
= | 0.5 A |
What direction is the current?
The change reduces the number of field lines passing through the loop into the page. The field generated by the induced current must be into the page to oppose the change. That requires a clockwise current.
If we'd kept the magnetic field constant, what other ways could we have induced the same current (magnitude and direction) in the loop?
By changing the field we reduced the flux to zero over a 10 second period. Two other ways to do that are:
What if, instead of reducing the the field to zero in 10 seconds, we reduced it to zero in 2 seconds? Would anything change? Would anything stay the same?
The current would still be clockwise, but it would be 5 times as large. So, we'd get five times the current for 1/5 of the time. The product of current and time wuld be the same, but that represents the total charge. DQ = I Dt, so the same total charge flows every time.