Two straight parallel wires carry currents in opposite directions as shown in th

Question

Two straight parallel wires carry currents in opposite directions as shown in the figure. One of the wires carries a current of I2 = 11.5 A. Point A is the midpoint between the wires. The total distance between the wires is d = 12.2 cm. Point C is 5.82 cm to the right of the wire carrying current I2. Current I1 is adjusted so that the magnetic field at C is zero.

a) Calculate the value of the current I1. (in A)

b) Calculate the magnitude of the magnetic field at point A. (in T)

c) What is the force between two 2.83 m long segments of the wires? (in N)

Explanation / Answer

r = 0.0582 m

at point C

magnetic field due to I1, B1 = uo*I1/(2*pi*r)

magnetic field due to I2, B2 = uo*I2/(2*pi*(d+r))

at C net field = 0

B1 = B2

uo*I1/(2*pi*r) = uo*I2/(2*pi*(d+r))

I1/r = I2/(d+r)

I1/0.0582 = 11.5/(0.122+0.0582)

I1 = 3.714 A

++++++++++++++++++

(B)

at point A

magnetic field due to I1, B1 = uo*I1/(2*pi*d/2) (out of page)

magnetic field due to I2, B2 = uo*I2/(2*pi*d/2) (out of page)

Bnet = B1 + B2

Bnet = (uo/pid)*(I1+I2)

Bnet = 4*pi*10^-7*(11.5+3.714)/(pi*0.122)

Bnet = 4.99*10^-5 T

+++++++++++++++++++++++

c)

F = uo*I1*I2*L/(2*pi*d)

F = (4*pi*10^-7*11.5*3.714*2.83)/(2*pi*0.122)

F = 1.98*10^-4 N

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