Two long, parallel wires carry currents of I 1 = 3.15 A and I 2 = 5.15 A in the

Question

Two long, parallel wires carry currents of I1 = 3.15 A and I2 = 5.15 A in the directions indicated in the figure below, where d = 21.5 cm. Take the positive x direction to be to the right.

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

magnitude Two long, parallel wires carry currents of I1 = 3.15 A and I2 = 5.15 A in the directions indicated in the figure below, where d = 21.5 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 MuT direction degree counterclockwise from the +x axis (b) Find the magnitude and direction of the magnetic field at point P, located d = 21.5 cm above the wire carrying the 5.15-A current. magnitude MuT direction degree counterclockwise from the +x axis

Explanation / Answer

apply magnetic field B due to each wire as B = uoi/2pi R

so

at mid point between them, r = d/2 = 21.5/2 = 10.75 cm

so

Bnet in middle = uoi/2pir + uoi/2pi R

Bnet = 4*3.14 e-7 * (3.15 +5.15)/(2*3.14 * 0.1075)

Bnet = 1.544 *10^-5 tesla

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P is at a disatnce of 1.414d from I1 and d from I2

so B1 due to i1 is B1 = 4*3.14 e-7 * 3.15/(2*3.14 * 1.414* 0.215)

B1 = 2.07 uT

B2 =   4*3.14 e-7 * 5.15 /(2*3.14 *0.215)

B2 = 4.79 uT

so

Bnet^2 = B1^2 +B2^2   + 2 B1B2 cos theta

here theta = tan ^-1(4.79/2.07) =

theta = 66.6deg

so

Bnet at P ^2 = 2.07^2 + 4.79^2 + ( 2 * 2.07 * 4.79 * cos 66.6)

Bnet at P = 5.92 uT

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