The figure below shows two long, straight wires carrying currents I_1 = 1.80 A a
ID: 1601147 • Letter: T
Question
The figure below shows two long, straight wires carrying currents I_1 = 1.80 A and I_2 = 6.00 A into the page. The wires are separated by a distance a = 1.80 m. (a) What are the magnitude and direction of the magnetic field at point A, halfway between the two wires? magnitude mu T direction degree counterclockwise from the +x axis (b) What are the magnitude and direction of the magnetic field at point B, at a distance a = 1.80 m to the right of the top wire? magnitude mu T direction degree counterclockwise from the +x axisExplanation / Answer
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)
= (4pi x 10^-7 / (pi x 1.80) ) (6- 1.8)
= 0.93 x 10^-6 T
magnitude = 0.93 uT
along +ve x axis.
direction : angle = 0
b) B1 = (4pi x 10^-7 x 1.80) / (2 x pi x 1.80 ) (- j)
B1 = - 2.0 x 10^-7 j
@ = tan^-1(a/a) = 45 deg
B2 = [(4pi x 10^-7 x 1.80) / (2 x pi x sqrt(1.8^2 + 1.8^2) ) ] (cos45i - sin45j)
B2 = 1.0 x 10^-7 i - 1.0 x 10^-7 j T
B = B1 + B2 = 1.0 x 10^-7i - 3.0 x 10^-7 j
magnitude = sqrt(1.0^2 + 3.0^2) x 10^-7 = 3.16 x 10^-7 T
= 0.316 uT
direction = 360 - tan^-1(3.16 / 1.0) = 287.56 deg