11. -10 points SerCP11 19.P050. My NotesAsk Your Teacher Two long, parallel wire

Question

11. -10 points SerCP11 19.P050. My NotesAsk Your Teacher Two long, parallel wires carry currents of 11 = 3.24 A and 12 = 5.05 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 (d20.0 cm). magnitude direction from the positive 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 5.05-A current. magnitude direction to the left of the vertical Need Help? Read

Explanation / Answer

(A) due to I1, B1 = u0 I1 / (2 pi d/2) upward  

due to I2. B2 = u0 I2 / ( pi d) downward  

B = B2 - B1 = u0 (I2 - I1) / (pi d )

= (4pi x 10^-7)(5.05 - 3.24) / ( pi 0.20)

= 3.62 x 10^-6 T downward

magnitude = 3.62 x 10^-6 T Or 3.62 uT

direction = 90 deg

(B) due to I2, B2 = (u0 I2 / 2 pi d )(-i)

= - 5.05 x 10^-6 i

due to I1, B1 = (u0 I1 / 2 pi d)( -cos45i + sin45j)

B1 = - 2.29 x 10^-6 i + 2.29 x 10^-6 j

B = B1 + B2 = - 7.34 x 10^-6 i + 2.29 x 10^-6 j

magnitude = sqrt(7.34^2 + 2.29^2) x 10^-6

= 7.69 x 10^-6 T Or 7.69 uT

direction = tan^-1(7.34/2.29) = 72.7 deg

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