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Samuelb88

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## 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*

**L**x

**B**(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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