Two insulated wires perpendicular to each other in the same plane carry currents

Question

Two insulated wires perpendicular to each other in the same plane carry currents as shown in the figure (Figure 1) . Assume that I = 11 A and d = 15 cm .

Part A

Find the magnitude of the net magnetic field these wires produce at points P and Q if the 10 A current is to the right.

Express your answers using two significant figures separated by a comma.

Part B

Find the magnitude of the net magnetic field these wires produce at points P and Q if the 10 A current is to the left.

Express your answers using two significant figures separated by a comma.

Explanation / Answer

part(A)

at P

magnetic field due to vertical wire = B1 = uo*I/(2*pi*d)

B1 = (4*3.14*10^-7*11)/(2*3.14*0.15) = 1.47*10^-5 T into the page

magnetic field due to horizantal wire = B2 = (uo*I2)/(2*pi*r)

r = 0.08 m

I2 = 10 A

B2 = (4*3.14*10^-7*10)/(2*3.14*0.08) = 2.5*10^-5 T in to the page

BP = B1 + B2 = 3.97*10^-5 T

at Q

magnetic field due to vertical wire = B1 = uo*I/(2*pi*d)

B1 = (4*3.14*10^-7*11)/(2*3.14*0.15) = 1.47*10^-5 T out of the page

magnetic field due to horizantal wire = B2 = (uo*I2)/(2*pi*r)

r = 0.08 m

I2 = 10 A

B2 = (4*3.14*10^-7*10)/(2*3.14*0.08) = 2.5*10^-5 T out of the page

BQ = B1 + B2 = 3.97*10^-5 T

part(B)

at P

magnetic field due to vertical wire = B1 = uo*I/(2*pi*d)

B1 = (4*3.14*10^-7*11)/(2*3.14*0.15) = 1.47*10^-5 T into the page

magnetic field due to horizantal wire = B2 = (uo*I2)/(2*pi*r)

r = 0.08 m

I2 = 10 A

B2 = (4*3.14*10^-7*10)/(2*3.14*0.08) = 2.5*10^-5 T out of the page

BP = B1 + B2 = 1.03*10^-5 T

at Q

magnetic field due to vertical wire = B1 = uo*I/(2*pi*d)

B1 = (4*3.14*10^-7*11)/(2*3.14*0.15) = 1.47*10^-5 T out of the page

magnetic field due to horizantal wire = B2 = (uo*I2)/(2*pi*r)

r = 0.08 m

I2 = 10 A

B2 = (4*3.14*10^-7*10)/(2*3.14*0.08) = 2.5*10^-5 T into the page

BQ = B1 - B2 = 1.03*10^-5 T

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