Two long, parallel wires carry currents of I 1 = 2.60 A and I 2 = 5.35 A in the

Question

Two long, parallel wires carry currents of I1 = 2.60 A and I2 = 5.35 A in the directions indicated in the figure below, where d = 19.0 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.

(b) Find the magnitude and direction of the magnetic field at point P, located d = 19.0 cm above the wire carrying the 5.35-A current.

Explanation / Answer

A) at a point midway between the wires is B = B1-B2

B1 = mu_0*i1/(2*pi*r) = (4*3.142*10^-7*2.6)/(2*3.142*0.095) = 5.47*10^-6 T

B2 = mu_o*i2/(2*pi*r) = (4*3.142*10^-7*5.35)/(2*3.142*0.095) = 11.2*10^-6 T

then B = (5.47-11.2)*10^-6 = -5.73*10^-6

Answer is magnitude 5.73 uT
270 degrees counterclockwise from the +X-axis
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Bx = -(mu_0/(2*pi))((i2/d)-(i1*cos(45)/1.414d))

Bx = -(2*10^-7)*((5.35/0.19)-(2.6*cos(45)/(1.414*0.19))

Bx = -4.26*10^-6

By = mu_0*i1*sin(45)/(2*3.142*1.414*0.19) = (4*3.142*10^-7*2.6*sin(45))/(2*3.142*1.414*0.19) = 1.36*10^-6 T

B = sqrt(Bx^2+By^2) = sqrt(4.26^2+1.36^2) = 4.47*10^-6 T

theta = atan(By/Bx) = atan(-1.36/4.26) = -17.7 degrees

Answers is magnitude is 4.47*10^-6 T

theta = 162.3 degrees counterclockwise from the _x-axis

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