Two long, parallel wires carry currents of I 1 = 2.88 A and I 2 = 4.60 A in the

Question

Two long, parallel wires carry currents of I1 = 2.88 A and I2 = 4.60 A in the direction indicated in the figure below. (Choose the line running from wire 1 to wire 2 as the positive x-axis and the line running upward from wire 1 as the positive y-axis.)

(a) Find the magnitude and direction of the magnetic field at a point midway between the wires (d = 20.0 cm).

(b) Find the magnitude and direction of the magnetic field at point P, located d = 20.0 cm above the wire carrying the 4.60-A current.

Explanation / Answer

b)

here

B1 = u0 * I / (2*pie) * r

r = sqrt(20^2 + 20^2) = 28.3 cm = 0.283 m

B1 = (4 * 3.14 * 10^-7 * 2.88 ) / ( 2 * 3.14 * 0.283)

B1 = 2.035 uT

then by using the same formula

B2 = (4 * pie * 10^-7 * 4.6) / ( 2 * 3.14 * 0.283)

B2 = 3.25 uT

then

Bx = 4.6 uT + 2.035 uT * cos(45deg) = 6.04 uT

By = 2.035 uT * sin(45deg) = 1.44 uT

B = sqrt( Bx^2 + By^2 )

B = sqrt( 6.04^2 + 1.44^2 ) = 6.21 uT

theta = tan^-1(1.44 / 6.04) = 13.41 deg

a)

B1 = (4 * 3.14 * 10^-7 * 2.88 ) / ( 2 * 3.14 * 0.1)

B1 = 5.76 uT

B2 = (4 * pie * 10^-7 * 4.6) / ( 2 * pie * 0.1)

B2 = 9.2 uT

net B = 9.2 - 5.76 = 3.44 uT

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