28.22 Two long, parallel transmission lines, 38.0 cmapart, carry 28.0-A and 76.0

Question

28.22

Two long, parallel transmission lines, 38.0 cmapart, carry 28.0-A and 76.0-A currents.

Part A

Find all locations where the net magnetic field of the two wires is zero if these currents are in the same direction.

Assume that the positive  x axis is directed from the 76.0-A wire to the 28.0-A wire perpendicular to the wires, with the origin on the 76.0-A wire.

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Part B

Find all locations where the net magnetic field of the two wires is zero if these currents are in the opposite direction.

Assume that the positive  x axis is directed from the 76.0-A wire to the 28.0-A wire perpendicular to the wires, with the origin on the 76.0-A wire.

28.22

Two long, parallel transmission lines, 38.0 cmapart, carry 28.0-A and 76.0-A currents.

Part A

Find all locations where the net magnetic field of the two wires is zero if these currents are in the same direction.

Assume that the positive  x axis is directed from the 76.0-A wire to the 28.0-A wire perpendicular to the wires, with the origin on the 76.0-A wire.

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Part B

Find all locations where the net magnetic field of the two wires is zero if these currents are in the opposite direction.

Assume that the positive  x axis is directed from the 76.0-A wire to the 28.0-A wire perpendicular to the wires, with the origin on the 76.0-A wire.

Explanation / Answer

if the currents are in the same direction then ,in between the wires the field is zero

let at a distance x from 76 A the field is zero

then Field due to 76 A = Field due to 28 A

B1 = B2

mu_o*i1/(2*pi*x) = mu_o*i2/(2*pi*(0.38-x))

i1/x = i2/(0.38-x)

76/x = 28/(0.38-x)

x = 0.277 m = 27.7 cm

Part B) in between the wires the field due to two wires is in same direction ,So net field never be zero in between the wires

So out side the wires ,i.e at a distance x to the left of the 76 A current carrying wire the field is zero

then

field due to 76 A = Field due to 28 A

i1/x = i2/(0.38+x)

76/x = 28/(0.38+x)

x = 0.601 m to the left of the 76 A wire

ans also at a distance x to the right of the 76 A wire ,the field is zero

then

i1/x = i2/(x-0.38)

76/x = 28/(x-0.38)

x = 0.601 m to the right of the 76 A wire

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