Question

Three long, straight wires all exist in the xy-plane. One carries a current of 3.30 AA in the positive y-direction through the point x=-3.40 mm . The second carries a current of 19.8 AA in the positive y-direction through the point x=9.00 mm . The third wire carries a current in the negative y-direction and goes through the point x=4.00 mm . What current does the third wire carry if the magnetic field at the position x=1.25 mm is exactly zero?

Answer 1

Magnetic field due to long current carrying wire is given by,

B = *I/(2*pi*d)

given, magnetic field at point x=1.25 m is equals to zero.

then, B1 + B2 + B3 = 0

here, B1 = magnetic field due to wire 1 = *I1/(2*pi*d1) (towards -z axis)

B2 = magnetic field due to wire 2 = *I2/(2*pi*d2) (towards +z axis)

B3 = magnetic field due to wire 3 = *I3/(2*pi*d3) (towards -z axis)

given, I1 = 3.30 A, I2 = 19.8 A

= 4*pi*10^-7

d1 = distance of wire 1 from 'x=1.25 m' = 1.25 + 3.40 = 4.65 m

d2 = distance of wire 2 from 'x=1.25 m' = 9.00 - 1.25 = 7.75 m

d3 = distance of wire 3 from 'x=1.25 m' = 4.00 - 1.25 = 2.75 m

then, -*I1/(2*pi*d1) + *I2/(2*pi*d2) - *I3/(2*pi*d3) = 0

[/(2*pi)]*(-I1/d1 + I2/d2 - I3/d3) = 0

-3.30/4.65 + 19.8/7.75 - I3/2.75 = 0

I3 = 2.75*(-3.30/4.65 + 19.8/7.75)

I3 = 5.07 A

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