Simple question about direction of force due to current

In summary: So if the magnetic field strength is decreasing, then the magnetic flux should be decreasing as well and the time-rate of change of the magnetic flux would be "negative". 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.
  • #1
Samuelb88
162
0

Homework Statement


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"

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.

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. [tex]w = (\frac{1}{LC}\right) )^(^1^/^2^) = 646. (rad./s)[/tex]

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

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

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

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
 
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  • #2
(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.
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
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 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.

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
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
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:

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

And evaluating I(.0042):

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

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.
 

1. What is the direction of force due to a current?

The direction of force due to a current is determined by the right-hand rule. If you point your right thumb in the direction of the current, the direction of the force will be perpendicular to your fingers, in the direction of your curled fingers.

2. Does the direction of the current affect the direction of the force?

Yes, the direction of the current does affect the direction of the force. The direction of the force will always be perpendicular to the direction of the current, as determined by the right-hand rule.

3. How does the strength of the current affect the force?

The strength of the current directly affects the force, as the force is directly proportional to the strength of the current. This means that a stronger current will result in a stronger force, and vice versa.

4. Can the direction of the force be reversed?

Yes, the direction of the force can be reversed by changing the direction of the current. If the direction of the current is reversed, the direction of the force will also be reversed, as determined by the right-hand rule.

5. Is the direction of the force the same in all parts of a circuit?

No, the direction of the force can vary in different parts of a circuit. This is because the direction of the current can change as it passes through different components, and the direction of the force is always perpendicular to the current.

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