Two long, parallel wires carry currents of I1 = 3.16 A and I2 = 5.45 A in the di

Question

Two long, parallel wires carry currents of I1 = 3.16 A and I2 = 5.45 A in the direction indicated in the figure below. (Choose the line running from wire 1 to wire 2 as the positive x-direction and the line running upward as the positive y-direction.)

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

magnitude _______ ?T

direction_______ ? (counterclockwise from the +x-axis)

(b) Find the magnitude and direction of the magnetic field at point P, located d = 20.0 cm above the wire carrying the 5.45-A current.

magnitude_______ ?T

direction________ ? (counterclockwise from the +x-axis)

Explanation / Answer

Use right hand rule to determine the direction of the two magnetic fields at the point, so they will be perpendicular and if we name them B1 and B2 the magnitude of their vector sum will be adopted by the formula B = v(B1² + B2²) Now we shall calculate each of B1 and B2 by means of the formula B = (µ0/2p)(I/r) where µ0 is 4p*10^-7T.m/A , I is the current within the wire and r is the distance of the point from the wire. Doing use of the formula we can derive any of B1 and B2 so we place them in B = v(B1² + B2²) and the out is what we want. I grant the calcualtions to you! Hope this helped.

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