Two very long wires both carry current toward the west One is directly above the
ID: 1455367 • Letter: T
Question
Two very long wires both carry current toward the west One is directly above the other, and they are 2.0 cm apart. The top wire (hanging by a series of insulating support cables) carries a current of 300 A. The bottom, resting on top of a long insulating table, carries a current of 100 A. 10) What is the magnetic field at a point 2.0 cm north of the top wire due to: a) the top wire only? 3.0 mT [D] b) the bottom wire only? 0.707 mT[N45degreeD] c) both wires? 3.54 mT[N82 degreeD] 11) What is the total magnetic field at a point 2.0 cm north of the bottom wire? 2.92 mT[D31 degreeS] 12) What is the magnetic field halfway between the wires? 4.0 mT [south] 13) What is the magnetic field sensed by the bottom wire? 3.0 mT [south] 14) What magnetic force (per unit length) acts on the bottom wire? 0.30 N/m [up] 15) If there is no normal force acting on the bottom wire, what must its mass (per unit length) be? 0.0306 kg/m 16) What is the magnetic field sensed by the top wire? 1.0 mT[nouth] 17) What magnetic force (per unit length) acts on the top wire? Does your answer agree with Newton's Third Law? 0.30 N/m [down]Explanation / Answer
Hi,
In this case, as the wires are said to be very long, we can approximate the wires to be infinitely long. If we do that, then we can use the following equation to find the magnetic field produced by any wire in any point of the space:
B = (uoI)/(2r) ; where B is the magnetic field, I is the current in the wire, r is the distance from the wire and uo is the vacuum permeability, which value is 4*10-7 N/A2.
Besides the force (per unit of length) felt by any wire due to the action of the magnetic field created by the other can be calculated as follows:
(Fb/L) = (uo I1 I2)/(2d) ; where (Fb/L) is the magnetic force per unit of length, I1 and I2 are the currents of the wires and d is the distance between them.
Also, if the two wires have currents flowing to the same direction, then they will attract each other, otherwise they will repel each other.
On a side note, to know the direction of the magnetic field one can solve the vectorial product I x r ; where I is a vector that points in the same direction that the current flow and r is a vector that points towards the point where we want to find the magnetic field.
Finally, it important to remember that vectors can be added just like numbers, so the effect of both wires over the same point can be obtained adding the individual effect caused by each wire.
(10) Top wire: r = 2 cm ; it goes down (we know this thanks to the right hand rule)
B = (4*10-7 N/A2)*(300 A)/(2*2*10-2m) = 3*10-3 T
Vector: B = 3*10-3 T (-j)
Bottom wire: r = 2.83 cm; it has a component of equal value on each axis (down and north)
B = (4*10-7 N/A2)*(100 A)/(2*2.83*10-2m) = 7.07*10-4 T
Bx = cos(45°) B = 4.999*10-4 T ::::::::: By = Bx = 4.999*10-4 T
Vector: B = [ 4.99*10-4(i) ; 4.99*10-4(-j) ] T
Both wires: Bx = 7.07*10-4 T (i) ; By = [ 3*10-3 T + 4.999*10-4 T ] (-j) = 3.5*10-3 T (-j)
Vector: B = [ 4.999*10-4(i) ; 3.5*10-3 (-j) ] T :::::::::: B = ( (4.999*10-4)2 + (3.5*10-3)2 )1/2 T = 3.54*10-3 T
(11) Top wire: r = 2.83 cm ; it has two components of equal size over the axis down and south.
B = (4*10-7 N/A2)*(300 A)/(2*2.83*10-2m) = 2.12*10-3 T
Bx = By = -cos(45°)B = -1.499*10-3 T
Bottom wire: r = 2 cm; it points downward.
B = (4*10-7 N/A2)*(100 A)/(2*2*10-2m) = 1*10-3 T
Bx = 0 T ; By = -1*10-3
Both wires: Bx = -1.499*10-3 T ; By = (-1*10-3 - 1.499*10-3 )T = -2.499*10-3 T
Vector: B = [ -1.499*10-3 T i ; -2.499*10-3 T j ] ::::::::::: B = ( (-2.499*10-3)2 + (1.499*10-3)2 )1/2 T = 2.92*10-3 T
(12) In this case the upper wire produces a magnetic field directed to the south, while the other wire produces a magnetic field in the opposite direction. (d = 1 cm )
Bu = (4*10-7 N/A2)*(300 A)/(2*1*10-2m) = 6.0*10-3 T ; magnetic field produced by the upper wire
Bd = (4*10-7 N/A2)*(100 A)/(2*1*10-2m) = 2*10-3 T ; magnetic field produced by the buttom wire
B = Bu - Bd = 4*10-3 T ; as the upper wire had a bigger field, this field points towards south.
(13) This wire will feel the magnetic field produced by the top wire. So we simply have to calculate the magnetic field produced by the top wire at a point 2 cm below it. However, we have calculated the magnetic field produced 1 cm below said wire, so the magnetic field at 2 cm below should be the haft of that value.
(B2/Bu) = (1 cm/2 cm) :::::: B2 = 6.0*10-3 T (1/2) = 3*10-3 T
Of course, the direction is the same to the previous case, so this field points towards south.
(14) The magnetic force per unit of length in both wires is equal to:
(Fb/L) = (4*10-7 N/A2)*(100 A)*(300 A)/(2*2*10-2m) = 0.3 N/m
An of course they attract each other, so the force over the bottom wire will point upwards.
Note: up to this point you should be able to solve the rest of the questions. if you can't, please post another question.
I hope it helps.
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