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Problem 3: The foil shielding on a power cable, carrying a DC current I = 0.39 A, has broken off. The unshielded part of the wire is x 0.021 meters long. The shielded parts of the wire do not contribute to the magnetic field outside the wire. Point P is above the right-hand end of the unshielded section, a distance x/2 above the wire. The current flows to the right (in the positive direction) as shown. The small length of wire, dl, a distance l from the midpoint of the unshielded section of the wire, contributes a differential magnetic field dB at point P. Ignore any edge effects. Part (a) Input an expression for the magnitude of the differential magnetic field, dB, generated at point P by the current moving through the segment of wire dl in terms of given parameters and fundamental constants. SchematicChoice : Holdl Holldl Holxdl dB = dB dB 3 8nx (() + 12 2nx + 1² + Holxdl Holxdl dB dB %3D %D 8nl () + 1² ) 2 4п + 1? Part (b) Perform the indefinite integral from part a. SchematicChoice : Holdl Holxl Holxl B = B = B = 1/2 1/2 2mx () (21 + 27 () + 1²) 8πί Holl B = 1/2 2πχ + Part (c) Select the limits of integration that will correctly calculate the magnetic field at P due to the current in the unshielded length of wire. MultipleChoice : 1) l = -00 to l = 0 2) l = -x to l = x 3) 1 = 0 to l = 00 4) l = -x to l = 0 5) l = -00 to l = 0 6) 1 = -x/2 to l= x/2 ミニ

Problem 3: The foil shielding on a power cable, carrying a DC current I = 0.39 A, has broken off. The unshielded part of the wire is x 0.021 meters long. The shielded parts of the wire do not contribute to the magnetic field outside the wire. Point P is above the right-hand end of the unshielded section, a distance x/2 above the wire. The current flows to the right (in the positive direction) as shown. The small length of wire, dl, a distance l from the midpoint of the unshielded section of the wire, contributes a differential magnetic field dB at point P. Ignore any edge effects. Part (a) Input an expression for the magnitude of the differential magnetic field, dB, generated at point P by the current moving through the segment of wire dl in terms of given parameters and fundamental constants. SchematicChoice : Holdl Holldl Holxdl dB = dB dB 3 8nx (() + 12 2nx + 1² + Holxdl Holxdl dB dB %3D %D 8nl () + 1² ) 2 4п + 1? Part (b) Perform the indefinite integral from part a. SchematicChoice : Holdl Holxl Holxl B = B = B = 1/2 1/2 2mx () (21 + 27 () + 1²) 8πί Holl B = 1/2 2πχ + Part (c) Select the limits of integration that will correctly calculate the magnetic field at P due to the current in the unshielded length of wire. MultipleChoice : 1) l = -00 to l = 0 2) l = -x to l = x 3) 1 = 0 to l = 00 4) l = -x to l = 0 5) l = -00 to l = 0 6) 1 = -x/2 to l= x/2 ミニ

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Problem 3: The foil shielding on a power cable, carrying a DC current I = 0.39 A, has broken off. The unshielded part of the wire is x = 0.021 meters long. The shielded parts of the wire do not contribute to the magnetic field outside the wire. Point P is above the right-hand end of the unshielded section, a distance x/2 above the wire. The current flows to the right (in the positive direction) as shown. The small length of wire, dl, a distance l from the midpoint of the unshielded section of the wire, contributes a differential magnetic field dB at point P. Ignore any edge effects. Part (a) Input an expression for the magnitude of the differential magnetic field, dB, generated at point P by the current moving through the segment of wire dl in terms of given parameters and fundamental constants. SchematicChoice : Holdl Holldl Holxdl dB = dB dB 2mx () + 1² )° 8tx ((5) + 12 + 12 Holxdl Holxdl dB = dB (6)* + 2 ()- 8nl + 12 Part (b) Perform the indefinite integral from part a. SchematicChoice : Holdl Holxl Holxl B = 1/2 B = B = 1/2 2m () + 1) 2пх + [² 8nl + [2 Holl B = 1/2 2πχ Part (c) Select the limits of integration that will correctly calculate the magnetic field at P due to the current in the unshielded length of wire. MultipleChoice : 1) l = -00 to l = 0 2) 1 = -x to l = x 3) 1 = 0 to l = 00 4) l = -x to l = 0 5) 1 = -00 to l = 0 6) l = -x/2 to l = x/2 Part (d) Evaluate the expression derived in part (b) using the endpoints selected in part (c). Expression : B = Select from the variables below to write your expression. Note that all variables may not be required. a, B, µo, T, 0, d, g, h, I, j, k, m, P, t, x Part (e) Determine the strength of the magnetic field (in tesla) at point P. Numeric : A numeric value is expected and not an expression. B = Part (f) In what direction will the magnetic field point at point P due to the current in the unshielded portion of the wire? MultipleChoice : 1) 2) Left 3) Up 4) Right 5) Down 6) Out of the page 7) A direction not listed. 8) Into the page
Transcribed Image Text:Problem 3: The foil shielding on a power cable, carrying a DC current I = 0.39 A, has broken off. The unshielded part of the wire is x = 0.021 meters long. The shielded parts of the wire do not contribute to the magnetic field outside the wire. Point P is above the right-hand end of the unshielded section, a distance x/2 above the wire. The current flows to the right (in the positive direction) as shown. The small length of wire, dl, a distance l from the midpoint of the unshielded section of the wire, contributes a differential magnetic field dB at point P. Ignore any edge effects. Part (a) Input an expression for the magnitude of the differential magnetic field, dB, generated at point P by the current moving through the segment of wire dl in terms of given parameters and fundamental constants. SchematicChoice : Holdl Holldl Holxdl dB = dB dB 2mx () + 1² )° 8tx ((5) + 12 + 12 Holxdl Holxdl dB = dB (6)* + 2 ()- 8nl + 12 Part (b) Perform the indefinite integral from part a. SchematicChoice : Holdl Holxl Holxl B = 1/2 B = B = 1/2 2m () + 1) 2пх + [² 8nl + [2 Holl B = 1/2 2πχ Part (c) Select the limits of integration that will correctly calculate the magnetic field at P due to the current in the unshielded length of wire. MultipleChoice : 1) l = -00 to l = 0 2) 1 = -x to l = x 3) 1 = 0 to l = 00 4) l = -x to l = 0 5) 1 = -00 to l = 0 6) l = -x/2 to l = x/2 Part (d) Evaluate the expression derived in part (b) using the endpoints selected in part (c). Expression : B = Select from the variables below to write your expression. Note that all variables may not be required. a, B, µo, T, 0, d, g, h, I, j, k, m, P, t, x Part (e) Determine the strength of the magnetic field (in tesla) at point P. Numeric : A numeric value is expected and not an expression. B = Part (f) In what direction will the magnetic field point at point P due to the current in the unshielded portion of the wire? MultipleChoice : 1) 2) Left 3) Up 4) Right 5) Down 6) Out of the page 7) A direction not listed. 8) Into the page
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