Two parallel wires, separated by a distance of 20 cm, are carrying currents (not

Question

Two parallel wires, separated by a distance of 20 cm, are carrying currents (not necessarily the same). One wire carries a current of 12 A in the direction shown. Point C is a distance of 5 cm to the right of the 12 A current and Point A is halfway between the two wires. .

What is the magnitude of the B-field at Point C due to the 12 A current?

What is the direction of the B-Field at Point C due to the 12 A current?

If the current on the left is adjusted so that the total magnetic field (due to both wires) at Point C is zero, find the magnitude of the left hand current.

In which direction would the left hand current have to flow for the total magnetic field at Point C to be zero?

If the currents remain the same as found in above, what will be the magnitude of the magnetic field for Point A (remember, this will be the result of both wires (don’t forget to take account the directions of each B-field from each wire)?

What is the magnitude and direction of the B-Field due to the left hand current at the location of the right hand current?

What are the magnitude and direction of the force (remember, there is a formula for the force on a current-carrying wire immersed in a magnetic field) on the right hand current due to the left hand current (HINT: The right hand current is immersed in the B-field created by the Left hand current, and you have already found that B-field)? Assume the wires are 2 meters long.

If an proton was passing Point A moving upward at 1x105 m/s parallel to the wires, what force would that proton experience? (Remember, you already determined the B-field at Point A)

please explain and show how you get the answers

Explanation / Answer

1)

BRc = uo*I/(2*pi*r)

BRc = (4*pi*10^-7*12)/(2*pi*0.05) = 4.8*10^-5 A

2)

into the page

3)

BLc = BRc

uo*IL/(2*pi*(r+d)) = uo*IR/(2*pi*r)

IL/(5+20) = 12/5

IL = 60 A     <<----------answer

4)

direction

down wards <<----------answer

5)

BA = B1A + B2A = uo*IL/(2*pi*d/2) + uo*IR/(2*pi*d/2)

BA = (uo*(IL+IR))/(pid)

BA = (4*pi*10^-7*(60+12))/(pi*0.2) = 0.000144 T

+++++++++++++++

BL = uo*IL/(2*i*d)

BL = (4*pi*10^-7*60)/(2*pi*0.2) = 6*10^-5 T

++++++++++++++

FR = BL*l*IR = 6*10^-5*12*2 = 0.00144 N

+++++++++++++

F = BA*q*v = 0.000144*1.6*10^-19*1*10^5 = 2.304*10^-18 N

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