I know the answer for part A is 4.17x10^-5 N(^x). I also know the answer to part

Question

I know the answer for  part A is 4.17x10^-5 N(^x). I also know the answer to part C is 6x10^-7 N/m and that it is repelled. I do nto know how to solve these so I would like the answers and the work for all three parts. Make sure your answers match mine before you submit an answer for rating!

Two currents are arranged as shown, a: Find the magnitude of the force on a 5 mu C charge at point P if its velocity is 1 times 10 6 m/s. Use I1 = 2A, I 2 = 3A, D = 6m and L = 2m. Mark the direction of B rightarrow in regions I, II and III. If B rightarrow changes directions in those regions, make sure you mark that. Explain. 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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