1) Two straight parallel wires are separated by 7.6 cm . There is a 2.0-A curren

Question

1) Two straight parallel wires are separated by 7.6 cm . There is a 2.0-A current flowing in the first wire. Part A If the magnetic field strength is found to be zero between the two wires at a distance of 1.9 cm from the first wire, what is the magnitude of the current in the second wire? Express your answer using two significant figures and include the appropriate units. I = .......(value ) ........(units)
2) A thin 13-cm-long solenoid has a total of 410 turns of wire and carries a current of 1.5 A . Part A Calculate the field inside the solenoid near the center. Express your answer using two significant figures. B = __________ T

1) Two straight parallel wires are separated by 7.6 cm . There is a 2.0-A current flowing in the first wire. Part A If the magnetic field strength is found to be zero between the two wires at a distance of 1.9 cm from the first wire, what is the magnitude of the current in the second wire? Express your answer using two significant figures and include the appropriate units. I = .......(value ) ........(units)
2) A thin 13-cm-long solenoid has a total of 410 turns of wire and carries a current of 1.5 A . Part A Calculate the field inside the solenoid near the center. Express your answer using two significant figures. B = __________ T

Part A If the magnetic field strength is found to be zero between the two wires at a distance of 1.9 cm from the first wire, what is the magnitude of the current in the second wire? Express your answer using two significant figures and include the appropriate units. I = .......(value ) ........(units)
2) A thin 13-cm-long solenoid has a total of 410 turns of wire and carries a current of 1.5 A . Part A Calculate the field inside the solenoid near the center. Express your answer using two significant figures. B = __________ T

Explanation / Answer

If the magnetic field at 1.9cm is zero it means that at that distance the fields produced by the two wires cancel out.
At 1.9cm from the first wire the field strength is
B= uI/ 2pi r ... . where u is the permitivity of free space, 4.0 x 10^-7 N/A^2
B= (4.0 x 10^-7 N/A^2)( 2 A)/ 2(3.14)(0.019m)
B=( 8 x10^-7)/(0.019)
B= 6.704 x 10^-6 Tesla

The magnitude of the magnetic field from the other wire must be equal in magnitude at 2.7cm away (7.6- 1.9) and opposite in direction. Using the right hand rule (thumb points in direction of current and fingers curl around the wire to show the direction of the magnetic field) you can see that for the magnetic fields to be in opposite direction between the two wires the current must be flowing in the same direction as the first wire. Using the formula for magnetic field strength you can then determine the current needed.

6.704 x10^-6 T= (4.0 x 10^-7) I/ 2 (3.14)(0.057m)
THen solve for I

I=5.99 A

Therefore there is a current of 5.99A in the wire and it is flowing in the same direction as the first.B = Uo*I*(N/L)

Where :

Uo = 4*pi*10^-7

I = current

N = 410 turns

L = 13*10^-2 meters

now, replacing :

B = 4*pi*10^-7*1.5*410 / 13*^10-2


B = 5.944i*10^-3 (Tesla)

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