The figure below shows two long, straight wires carrying currents I1 = 1.70 A an
ID: 1601898 • Letter: T
Question
The figure below shows two long, straight wires carrying currents I1 = 1.70 A and I2 = 5.50 A into the page. The wires are separated by a distance a = 1.60 m.
(a) What are the magnitude and direction of the magnetic field at point A, halfway between the two wires? magnitude µT direction ° counterclockwise from the +x axis
(b) What are the magnitude and direction of the magnetic field at point B, at a distance a = 1.60 m to the right of the top wire? magnitude µT direction ° counterclockwise from the +x axis
a 8)11 B a x A 8117Explanation / Answer
Given; I1 = 1.70 A I2 = 5.50 A a = 1.60 m
(a) field due to current carrying wire
B =(u0*I ) / (2*pi*r)
at A:
B = - [ (u0*I1) / (pi*a ) ] + [ (u0*I2) / (pi*a) ]
= (u0 /pi*a) (I2 - I1)
= (4*pi*10^-7/ (pi*1.60)) *(5.50-1.70)
= 95*10^-6 T
magnitude = 95 micro T
along +ve x axis.
angel = 0 degree
(b) B1 = (4*pi*10^-7*1.70) / (2*pi*1.60 ) (- j)
B1 = - 2.125*10^-7 j
theta = tan^-1(a/a) = 45 degree
B2 = [(4*pi*10^-7*5.50) / (2*pi*sqrt(1.6^2 + 1.6^2) ) ] (cos45i - sin45j)
B2 = 3.437*10^-7 i - 3.437*10^-7 j T
B = B1 + B2 = 3.437*10^-7 i - 5.562*10^-7 j
magnitude = sqrt(3.437^2 + 5.562^2)*10^-7 = 6.538*10^-7 T
= 0.6538 micro T
direction = 360 - tan^-1(5.562 / 3.437) = 301.713 degree