In the wire shown in (Figure 1) segment B C is an arc of a circle with radius 30
ID: 1637295 • Letter: I
Question
In the wire shown in (Figure 1) segment BC is an arc of a circle with radius 30.0 cm, and point P is at the center of curvature of the arc. Segment DA is an arc of a circle with radius 20.0 cm, and point P is at its center of curvature. Segments CD and AB are straight lines of length 10.0 cm each.
Part A
Calculate the magnitude of the magnetic field at a point P due to a current 11.1 A in the wire.
Express your answer with the appropriate units
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Part B
What is the direction of magnetic field?
Figure 1 of 1
In the wire shown in (Figure 1) segment BC is an arc of a circle with radius 30.0 cm, and point P is at the center of curvature of the arc. Segment DA is an arc of a circle with radius 20.0 cm, and point P is at its center of curvature. Segments CD and AB are straight lines of length 10.0 cm each.
Part A
Calculate the magnitude of the magnetic field at a point P due to a current 11.1 A in the wire.
Express your answer with the appropriate units
B =SubmitMy AnswersGive Up
Part B
What is the direction of magnetic field?
What is the direction of magnetic field? out of the page into the pageFigure 1 of 1
C P° AExplanation / Answer
Part (A)
magnetic field due to an arc of circular current carrying
wire = B = uo*i*theta / 4*pi*R
theta is in radians
magnetic field due to the arc BC
B1 = uo*i*(2*pi/3)/(4*pi*R1)
B1 = uo*i*2/(4*3*R1)
B1 = (4*3.14*10^-7*11.1*2)/(4*3*0.3)
= 7.7*10^-6 T out of the page
magnetic field due to the arc AD
B2 = uo*i*(2*pi/3)/(4*pi*R2)
B2 = uo*i*2/(4*3*R2)
B2 = (4*3.14*10^-7*11.1*2)/(4*3*0.2)
= 1.16*10^-5 T into the page
the line joinig the point P and segement CD and BC makes an angle 0
magnetic field due to segments CB and DA = B = 0
net magnetic field at P = Bnet = B1 + B2 = 3.9*10^-6 T into the page
Part (B)
The direction of magnetic field is into the page