Phys: Induction and Induced Current (2 loops)

[mscoder610]

I've been having trouble with the solution my book (and the solution manual) gives for a problem.

Basically, there are two inductors (wire coils): one is on the left, hooked to a battery and a switch. Then there's one to the right of it, hooked to only a resistor. The left switch is closed for a while, and then opened. The problem is to determine the direction of the current in resistor "ab" when the switch is opened (after being closed for a while).

That's it - the book and solution manual say the current goes from A to B (left to right) through the resistor, but it seems to me like it's the other way around.

I include a crudely drawn figure in Paint for reference:
http://www.malcolm-s.net/images/phys1.gif

This is a pretty simple problem, I'm just not understanding the answer given. Thanks for any help.

 

[marlon]

I've been having trouble with the solution my book (and the solution manual) gives for a problem.

Basically, there are two inductors (wire coils): one is on the left, hooked to a battery and a switch. Then there's one to the right of it, hooked to only a resistor. The left switch is closed for a while, and then opened. The problem is to determine the direction of the current in resistor "ab" when the switch is opened (after being closed for a while).

That's it - the book and solution manual say the current goes from A to B (left to right) through the resistor, but it seems to me like it's the other way around.

I include a crudely drawn figure in Paint for reference:
http://www.malcolm-s.net/images/phys1.gif

This is a pretty simple problem, I'm just not understanding the answer given. Thanks for any help.

Well actually, this picture sucks. i mean, for starters the two chains need to be interconnected by some ferromagnetic material (PLACED IN THE SOLENOIDS OF EACH CHAIN) in order to bring over the influence of the current (ie the generated B-field in the first solenoid) from the primary (with battery) to the secondary chain.

If the first chain is closed, the current flows and the B-field is generated in the solenoid. You have a varying magnetic flux and thus a current will be generated of which the associated magnetic field will oppose the primary B-field. However it is NOT this indiced current that is transported over the ferromagnetic material. This material get's magnetized in the same direction as the induced B-field. The result will be the 'birth' of a current in the secondary chain because of the varying B-field in the ferromagnetic material and therefore, because of the varying B-field in the solenoid of the second chain. The direction of the current will be opposite to the current in the FIRST chain because the induced current needs to oppose the B-field in the solenoid.

If you put an Ampere meter inside the second chain, it will go from 0 to one side and then back to zero when the current in the first chain is maximal (because then, the B-field transported by the ferromagnet will be maximal and CONSTANT.)

Now, let's open the first chain so that the current goes from the maximal value to zero. The B-field in the solenoid of the second chain will dimish and the Ampere meter will go from zero to the OPPOSITE side and then back to zero if the B-field has become ZERO. The induced current will be in the same direction of the current in the first chain because the associated generated B-field must oppose the loss in the solenoid's magnetic field that will diminish because the first chain is opened.

marlon

 

[mscoder610]

The induced current will be in the same direction of the current in the first chain because the associated generated B-field must oppose the loss in the solenoid's magnetic field that will diminish because the first chain is opened.

Wouldn't that be clockwise for both, and therefore B->A? That's how I understood it.
Thanks.