# Simple question about direction of force due to current

1. Mar 20, 2010

### Samuelb88

1. The problem statement, all variables and given/known data
A small loop is located near a LC circuit (see link). Given

q_0 = -50. uC
I_0 = -30. mA
L = 40. mH
C = 60. uF

http://yfrog.com/jlphysj" [Broken]

At t = 4.20 ms, determine (A) the sense of the induced current in the small loop, and (B) the direction of the net force on the small loop due to the current-current interaction.

3. The attempt at a solution

Before determining the sense of the induced current in the small loop... I determined an explicit expression for the charge function q(t) = Qcos[wt+P], that means finding the phase angle P, the maximum charge Q, and the angular frequency of oscillation w.

I. $$w = (\frac{1}{LC}\right) )^(^1^/^2^) = 646. (rad./s)$$

II. $$E_t_o_t = \frac{q_0^2}{2C}\right) + \frac{1}{2}LI_0^2 = 3.88*10^-5 J$$

So: $$Q = (2CE_t_o_t)^(^1^/^2^) = 6.82*10^-5 C$$

III. $$P = arccos(\frac{q_0}{Q}\right) ) = {+/-} 2.39 (rad.)$$

Since I_0 < 0, and I_0 = -wQsin[P] < 0, positive angle should be chosen, so P = + 2.39 (rad.)

(A) At t = .0042s, I(.0042) > 0, so this suggest current in LC circuit is flowing clockwise, as indicated by the arrow in the LC circuit indicating positive current flow. Using the right hand rule for current, I see the magnetic field at this time is directed out of the page in the small loop. So, using the right hand rule for current again, I see the sense of the induced current in the small loop is counterclockwise.

(B) Again, at t = .0042s, the direction of the net force on the small loop due to the current-current interaction seems like it should be directed downwards, that is, the force exerted on the small loop would be trying to "pull it closer" to the LC circuit, as suggested by the right hand rule when applied to the force vector equation F = I*LxB (the vector L is taken to be in the same direction as the induced current in the small loop).

Anyways, my professor has provided an answer sheet, and he has the direction of the net force on the small loop due to the current-current interaction is directed upwards (that is, away from the LC circuit) and I can't understand why. Could some please explain to me what I am not understanding about this? I'm very confused...

Thanks

- Sam

Last edited by a moderator: May 4, 2017
2. Mar 20, 2010

### JaWiB

Without reading too carefully (so you might still have the correct answer here), the induced current doesn't depend on the direction of the magnetic field, but the change in the magnetic flux. So if the magnetic field is increasing and you have determined its direction correctly, then you are correct. If the magnetic field is decreasing, then the current is in the opposite direction.

3. Mar 20, 2010

### Samuelb88

Ahh yes. At any rate, the answer sheet my professor has provided says the induced current is counterclockwise. Also, seeing as my argument is not valid, I guess that brings rise to another question. On his answer sheet, he has" the time-rate of change of the magnetic flux directed into the page, therefore the induced emf is counterclockwise -> induced current is counterclockwise."

So trying to understand this... (correct me if I am wrong) Calculating the time-rate of change of the current dI/dt and evaluating it at t=.0042s, it is seen that dI/dt < 0, so the current in the LC circuit must be decreasing. If the current in the LC circuit is decreasing, shouldn't the magnetic field strength be decreasing as well since in general, the magnetic field strength is proportional to the current? And if the magnetic field strength is decreasing, then the magnetic flux is decreasing and the time-rate of change of the magnetic flux is "negative" or... into the page because the field is directed out of the page (?) And so Lenz Law essentially says you should point your thumb in the opposite direction of the d(Phi)/dt and curl your fingers to determine the sense of the induced current. So by this, it would indicate the sense of the induced current in counterclockwise.

4. Mar 20, 2010

### JaWiB

You're right, I said that backwards (if the B field is out of the page and increasing, the induced current should be clockwise)

It does seem to me that there should be an attractive force between the loops. The only way I can see a repulsive force is if the current in the LC circuit is counter-clockwise and increasing. In that case, the magnetic field in the small loop is into the page and increasing, so the induced emf would create a moment out of the page and a current in the counter-clockwise direction (check me on this). Did your professor indicate the direction of the current in the LC circuit?

5. Mar 20, 2010

### Samuelb88

I probably should mention on his answer sheet (it is very messy and hard to follow) he has "t = 4.20 ms -> I = -8.9 mA, dI/dt = -28. A/s" However, I've evaluated the current function at t=.0042s multiple times, and unless I have forgot how to differentiate, then there is no way I(.0042) = -8.9 mA... At any rate, if the current in the LC circuit is "negative," then yes, that would mean its sense is counterclockwise. Also, since dI/dt < 0, this means the current in decreasing, and making the same argument as I made in my previous post {...}. It would seem like Lenz Law would now suggest the sense of the induced current is clockwise under the pretenses he has given on his answer sheet (that is, I < 0, dI/dt < 0). Uhhhhg...

Anyways, recalculating the current function one more time:

$$\frac{d(q(t))}{dt}\right) = I(t) = -wQsin[wt+P]$$

And evaluating I(.0042):

$$I(.0042) = -[646.][6.82*10^-^5]sin(646(.0042)+2.39) = .04074 A$$

I'm going to send you a link where you can view the paper I am looking at, as well as his answer sheet if you wish to look over it to perhaps get a better understanding... I understand if you don't want to work through it. At any rate, thanks for your help yo.