Question

A transmission wire oriented parallel to the x-axis carries a current of 200 A flowing along the +x direction. A second transmission wire also oriented parallel to the x-axis but lying at a perpendicular distance of 1.30 m below the first wire carries a current of 215 A flowing along the +x direction. What is the magnitude of the total magnetic field at a point 2.3 m directly below a point midway between the two wires?

What is the direction of the net magnetic field?

transmission wire oriented parallel to the x-axis carries a current of   200 A    flowing along the +x direction. A second transmission wire also oriented parallel to the x-axis but lying at a perpendicular distance of   1.30 m    below the first wire carries a current of   215 A     flowing along the +x direction. If the location where the net magnetic field is zero is at perpendicular distance  X from the first wire, what is X ?

A transmission wire oriented parallel to the x-axis carries a current of   200 A    flowing along the +x direction. A second transmission wire also oriented parallel to the x-axis but lying at a perpendicular distance of   1.30 m     below the first wire carries a current of   215 A    flowing along the -x direction. What is the magnitude of the total magnetic field at a point   2.3 m    directly above a point midway between the two wires?

Answer 1

(a)

Magnitude of the total magnetic field at a point 2.3 m directly below a point midway between the two wires,

Magnetic field due to upper wire,

B1 = u0*l1 / 2*pi*d1

d1 = (1.30 / 2) + 2.3 = 2.95 m

B1 = 4*pi*10^-7*200 / 2*pi*2.95 = 1.355*10^-5 T (into the page)

Magnetic field due to lower wire,

B2 = u0*l2 / 2*pi*d2

d2 = 2.3 - (1.30 / 2) = 1.65 m

B2 = 4*pi*10^-7*215 / 2*pi*1.65 = 2.606*10^-5 T (into the page)

Total magnetic field, B = B1 + B2

B = 3.96*10^-5 T

direction of the net magnetic field is into the page.

(b)

As given tha net magnetic field is zero is at perpendicular distance x from the first wire.

u0*l1 / 2*pi*x - u0*l2 / 2*pi*(1.30 - x ) = 0

200 / x = 215 / (1.30 - x)

415 x = 260

x = 0.626 m

(c)

Magnitude of the total magnetic field at a point 2.3 m directly above a point midway between the two wires,

Magnetic field due to upper wire,

B1 = u0*l1 / 2*pi*d1

d1 = 2.3 - (1.30 / 2) = 1.65 m

B1 = 4*pi*10^-7*200 / 2*pi*1.65 = 2.42*10^-5 T (upwards)

Magnetic field due to lower wire,

B2 = u0*l2 / 2*pi*d2

d2 = 2.3 + (1.30 / 2) = 2.95 m

B2 = 4*pi*10^-7*215 / 2*pi*2.95 = 1.457*10^-5 T (downwards)

Total magnetic fied

B = B1 - B2

B = 9.66*10^-6 T