The magnetic field is 1.50uT at a distance 40.0cm away from a long, straight wir

Question

The magnetic field is 1.50uT at a distance 40.0cm away from a long, straight wire. At one instant, the two conductors in a long household extension cord carry equal 2.02A currents in opposite directions. The two wires are 3.13mm apart.

Calculate the magnetic field 40.0cm way from the middle of the straight cord, in the plane of the two wires.

At what distance is it one-tenth as large?

The center wire in a coaxial cable carries current 2.02A in one direction, and the sheath around it carries current 2.02A in the opposite direction. What magnetic field does the cable create at points outside?

Explanation / Answer

a)

Given

R1-R2=3.13 mm

so

R1=0.4+(3.13*10^-3/2)

R1=0.401565 m

R2=0.4-(3.13*10^-3/2)

R2=0.398435 m

so Net Magnetic field is

Bnet=B2-B1 =(uoI/2pi)[1/R2 -1/R1]

Bnet =(uo*I/2pi)[R1-R2/R1R2]

Bnet=(4pi*10^-7*2.02/2pi)*(3.13*10^-3/0.401565*0.398435)

Bnet=7.9*10^-9 T or 7.9*10^-3 uT

b)

B=(1/10)*7.9*10^-9

B=7.9*10^-10 T

R1=R2=R (approx)

B=(uoI/2pi((R1-R2/R1R2)

7.9*10^-10=(4pi*10^-7*2.02/2pi)(3.13*10^-3/R^2)

R=1.6 m

c)

magnetic field at point outside the cable is

B=(uo/2piR)(I1-I2)

B=0 T

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