Two long thin parallel wires a distance d = 14.5 cm apart carry 25-A currents (I

Question

Two long thin parallel wires a distance d = 14.5 cm apart carry 25-A currents (I) in the same direction. The magnetic field is measured at point P, a distance d1 = 13.7 cm from the lower wire and a distance d2 = 6.7 cm from the upper wire.
a. Find the x and y components of the magnetic field at P due to the current in the lower wire only.

B1,x =

B1,y =

b. Find the x and y components of the magnetic field at P due to the current in the upper wire only.

B2,x =

B2,y =

c. Determine the magnitude |B| of the total magnetic field at P.

|B| =

Please include solution as well as answer. Thank you!

Explanation / Answer

B1 = 2*10^-7 * 25 / 0.137 = 3.65 * 10^-5 T

B1,x = - B1 sin(62.7 degree) = - 4.15 * 10^-5 T

B2,y = B1 cos(62.7 degree) = 2.14 * 10^-5 T

similarly,

B2= 7.46 * 10^-5 T

B2,x = B2 sin(20.31 degee) = 2.59 * 10^-5 T

B2,y = B2 cos(20.31 degee) = 7 * 10^-5 T

.'. lBl = [ sqrt { (2.59 - 4.15)^2 + (7 + 2.14)^2 } ] *10^-5 = 9.27 * 10^-5 T

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