Consider the long, straight, current-carrying wires shown below. One wire carrie
ID: 1784384 • Letter: C
Question
Consider the long, straight, current-carrying wires shown below. One wire carries a current of 7.6 A in the + y direction; the other wire carries a current of 4.5 A in the positive x direction.
(a) Conceptually: At which of the two labelled points do you expect the magnitude of the net magnetic field to be greater?
(A, B, or same at A&B)
(b) Calculate: What is the net magnetic field at each of these points?
(Do your answers match your predictions from part a?)
at point A:
size: __________ T, dir: (+x, -x, +y, -y, into the page, out of the page, field is zero)
at point B:
size: __________ T, dir: (+x, -x, +y, -y, into the page, out of the page, field is zero)
(c) What force would be felt by a proton placed at point B at rest?
size: __________ N, dir: (+x, -x, +y, -y, into the page, out of the page, field is zero)
(d) What force would be felt by a proton when passing through point A at 35 m/s in the +y dir?
size: __________ N, dir: (+x, -x, +y, -y, into the page, out of the page, field is zero)
Explanation / Answer
magnetic field at a perpendicular distance from curent carrying wire
B = uo*I/(2*pi*r)
(a)
net magnetic field at A > net magnetic field at B
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(b)
I1 = 76 A
I2 = 4.5 A
at point A
magnetic field due I1 is out of plane , B1z = uo*I1/(2*pi*r1)
B1z = 4*pi*10^-7*7.6/(2*pi*0.16) = 9.5*10^-6 T
magnetic field due I2 is out of plane , B2z = uo*I2/(2*pi*r1)
B2z = 4*pi*10^-7*4.5/(2*pi*0.16) = 5.625*10^-6 T
net magnetic field at A , B = B1z + B2z = 15.125 uT (out of page )
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at poin B
magnetic field due I1 is in to the plane , B1z = -uo*I1/(2*pi*r1)
B1z = -4*pi*10^-7*7.6/(2*pi*0.16) = -9.5*10^-6 T
magnetic field due I2 is out of plane , B2z = uo*I2/(2*pi*r1)
B2z = 4*pi*10^-7*4.5/(2*pi*0.16) = 5.625*10^-6 T
net magnetic field at A , B = B1z + B2z = -3.875 uT ( into the page)
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c)
magnetic force Fb = q*(v x B )
speed v = 0
magnetic force Fb = 0
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(d)
v = 35 j m/s
magnetic field B = 15.125 k
Fb = q*(vxB)
Fb = 1.6*10^-19*(35j x (15.125*10^-6 k) )
Fb = 1.6*10^-19*35*15.125*10^-6 i
Fb = 8.47*10^-23 N i
Fb = 8.47*10^-23 N ( +x axis )