Two long, parallel wires carry currents of I_1 = 2.75 A and I_2 = 4.85 A in the

Question

Two long, parallel wires carry currents of I_1 = 2.75 A and I_2 = 4.85 A in the directions indicated in the figure below, where d = 17.0 cm. (Take the positive x direction to be to the right.) Find the magnitude and direction of the magnetic field at a point midway between the wires, magnitude mu T direction counterclockwise from the +x axis find the magnitude and direction of the magnetic field at point P, located d - 17.0 cm above the wire carrying the 4.85 -A current, magnitude mu T direction counterclockwise from the +x axis

Explanation / Answer

a) at mid point,

due to I1,

B1 = (u0 I1 ) / ( 2 pi (d/2)) upward

= (4 pi x 10^-7 x 2.75 ) / (2 x pi x 0.17 / 2) = 6.47 x 10^-6 T upward

due to I2.

B2 = (4 pi x 10^-7 x 4.85 ) / (2 x pi x 0.17 / 2) = 11.41 x 10^-6 T downward

B = B1 - B2 = - 4.94 x 10^-6 T

(minus means downward )

B = 4.94 x 10^-6 T = 4.94 uT

angle = 270 deg

b) B2 = (4 pi x 10^-7 x 4.85 ) / (2 x pi x 0.17 ) to the left

B2 = 5.706 x 10^-6 T to the left

B2 = - 5.706i x 10^-6 T

field due to I1 will make 180 - 45 = 135 deg to +ve x axis.

B1 = [ (4 x pi x 10^-7 x 2.75) / (2 x pi x sqrt(0.17^2 + 0.17^2)] (cos135i + sin135j)

B1 = ( -1.618i + 1.618j ) x 10^-7 T

Bnet = B1 + B2

= (- 7.324i + 1.618j) x 10^-6

magnitude = sqrt[ 7.324^2 + 1.618^2] = 7.5 uT

angle = 180 + tan^-1(1.1618 / 7.324) = 167.5 deg

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