The two wires shown in the figure below are separated by d = 12.2 cm and carry c

Question

The two wires shown in the figure below are separated by d = 12.2 cm and carry currents of I = 5.25 A in opposite directions.

(a) Find the magnitude and direction of the net magnetic field at a point midway between the wires.

(b) Find the magnitude and direction of the net magnetic field at point P1, 12.2 cm to the right of the wire on the right.

(c) Find the magnitude and direction of the net magnetic field at point P2, 2d = 24.4 cm to the left of the wire on the left.

magnitude     µT direction     ---Select--- into the page out of the page toward the top of the page toward the bottom of the page toward the left side of the page toward the right side of the page no direction

Explanation / Answer

a)

magnetic field due to current through a wire is

B = uo*I/(2*pi*r)

r = distance from wire

uo = permeability constant = 4*pi*10^(-7) N/A^2

magnetic field at a point midway between the wires due to one wire

B1 = 4*pi*10^(-7)*5.25/(2*pi*(0.061)) = 17.2*10^(-6) = 17.2 uT

direction is into the page (according to right hand thumb rule)

magnetic field at a point midway between the wires due to another wire

B2 = 4*pi*10^(-7)*5.25/(2*pi*(0.061)) = 17.2*10^(-6) = 17.2 uT

direction is into the page (according to right hand thumb rule)

so the net magnetic field at a point midway between the wires

B = 17.2 uT + 17.2 uT = 34.4 uT

b)

magnetic field at p1 due to left wire

B1 = 4*pi*10^(-7)*5.25/(2*pi*(0.122 + 0.122)) = 4.3*10^(-6) = 4.3 uT

direction is into the page (according to right hand thumb rule)

magnetic field at p1 due to right wire

B2 = 4*pi*10^(-7)*5.25/(2*pi*(0.122)) = 8.6*10^(-6) = 8.6 uT

direction is out of the page (according to right hand thumb rule)

net magnetic field at p1 = B2 - B1 = 8.6 - 4.3 = 4.3 uT

and direction is out of the page

c)

magnetic field at p2 due to left wire

B1 = 4*pi*10^(-7)*5.25/(2*pi*(2*0.122)) = 4.3*10^(-6) = 4.3 uT

direction is out of the page (according to right hand thumb rule)

magnetic field at p2 due to right wire

B2 = 4*pi*10^(-7)*5.25/(2*pi*(3*0.122)) = 2.86*10^(-6) = 2.86 uT

direction is into the page (according to right hand thumb rule)

net magnetic field at p1 = B1 - B2 = 4.3 - 2.86 = 1.44 uT

and direction is out of the page

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