Two kong, parallel wires carry currents of lg 3.12 A and I2-4.55 A in the direct

Question

Two kong, parallel wires carry currents of lg 3.12 A and I2-4.55 A in the direction indicated in the running upward from wire 1 as the positive y axis.) figure below. (Choose the line running from wire 1 to wire 2 as the positive x-axis and the (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 magnetkc fieid at point P, located d-20.0 cm above the wire carrying the 4.55-A current. Need Help? EPR Submit Anewer Save Progress to the left of the vertical rtain s perconducting magnet intheform of a solenoid of length 0 34 man generate a magnetic neid of 8 5 T in its core when its cois carry a current of SO A. The winding, made 3 5 6 7 8

Explanation / Answer

Given,

I1 = 3.12 a ; I2 = 4.55 A ; d = 20 cm

a)The magnetic field at midpoint will be;

B = B1 - B2

B = u0 I1/2 pi r1 - u0 I2/2 pi r2

r1 = r2 = 20cm/2 = 10 cm = 0.1 m

B = u0/2 pi r(I1 - I2)

B = 4 pi x 10^-7/ 2 pi x 0.1 [ 3.12 - 4.55]

B = -2.86 x 10^-6 T

Hence, B = 2.86 x 10^-6 = 2.86 uT towards the bottom

b)The B field along X will be

Bx = B2 cos0 + B1 cos45

Bx = u0 I2/2 pi r2 + u0 I/2 pi r1 cos45

Bx = u0/ 2 pi (I2/r2 + I1 cos45/r1)

r1 = sqrt (20^2 + 20^2) = 28.3 cm = 0.283 m

Bx = 4 pi x 10^-7/ 2 pi (4.55/0.2 + 3.12 x .707/.283) = 6.11 x 10^-6 T

By = u0 I1/2 pi r1 sin(theta)

By = 4 pi x 10^-7 x 3.12 x .707/2 x pi x .283 = 1.56 x 10^-6 T

B = sqrt (6.11^2 + 1.56^2) x 10^-6 = 6.31 x 10^-6 T

theta = tan^-1(6.11/1.56) = 75.68 deh

Hence, B = 6.31 x 10^-6 T and theta = 75.68 deg

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