Two long, parallel wires carry currents of I 1 = 2.70 A and I 2 = 4.60 A in the

Question

Two long, parallel wires carry currents of I1 = 2.70 A and I2 = 4.60 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.)

Direction __90°__ 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 4.60-A current.

  
_______(Your response differs from the correct answer by more than 10%. Double check your calculations.) ° to the left of the vertical

PLEASE HELP! PLEASE MAKE SURE ITS CLEAR AND ANSWERS ARE WELL SHOWN! THANK YOU!

a) magnitude _______ Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. T  

Direction __90°__ 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 4.60-A current.

magnitude       
_______(Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.) T direction

  
_______(Your response differs from the correct answer by more than 10%. Double check your calculations.) ° to the left of the vertical

PLEASE HELP! PLEASE MAKE SURE ITS CLEAR AND ANSWERS ARE WELL SHOWN! THANK YOU!

P 2 d d

Explanation / Answer

Magnetic field due to a wire is proportional to I/R. [B = oI / 2R]

So, setting B1 = B2 and canceling constants,

we get I1 / R1 = I2 / R2 20(R2) = 5(R1) R1 = 4(R1) So, it should be 4 times further from the wire carrying 20A, or 4/5 of the distance separating the wires.

r = .8 * .20m = .16m (from the 20A wire)

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