a. Two currents are arranged as shown: Fin the magnitude of the force on a 5uC c

Question

a. Two currents are arranged as shown: Fin the magnitude of the force on a 5uC charge at point P if its velocity is 1x10^6m/s. Use I1=2A, I2= 3A, D= 6m, and L=2m.

b. Mark the direction of B in regions I, II, and III. If B changes directions in those regions, make sure you mark that. Explain.

c. Determine the force per unit length that the currents exert on each other. Are they attracted to each other or repelled? Explain.

a. Two currents are arranged as shown: Fin the magnitude of the force on a 5uC charge at point P if its velocity is 1x10^6m/s. Use I1=2A, I2= 3A, D= 6m, and L=2m. b. Mark the direction of B in regions I, II, and III. If B changes directions in those regions, make sure you mark that. Explain. c. Determine the force per unit length that the currents exert on each other. Are they attracted to each other or repelled? Explain.

Explanation / Answer

Part A)
The B field from wire 1 at the location of the charge is found from

B = uI/2pi(r)

B = (4pi X 10^-7)(2)/2pi(6)

B = 6.67 X 10^-8 T out of the page

The B field from wire 1 at the location of the charge is found from

B = uI/2pi(r)

B = (4pi X 10^-7)(3)/2pi(8)

B = 7.5 X 10^-8 T into the page

The net B = 7.5 X 10^-8 - 6.67 X 10^-8 = 8.3 X 10^-9 T

Then apply F = qvB

F = (5 X 10^-6)(1 X 10^6)(8.3 X 10^-9)

F = 4.17 X 10^-8 N

Part B)
In region I, by the right hand rule, and the formula applied above the B field is into the page closer to I1 and turns to out of the page the farther you get away.

I/r = I/r

2/r = 3/r + 2

2r + 4 = 3r

r = 4m. The switch happens 4 m from I1

In region 2, the B field is always into the page by the right hand rule

In region 3, the B field is always out of the page by the right hand rule and since I2 is greater than I1

Part C)
F/l = uII/2pir

F/l = (4pi X 10^-7)(2)(3)/(2pi)(2)

F/l = 6 X 10^-7 N/m and they repel since the currents are in opposite directions

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