The magnetic field 36.0 cm away from a long, straight wire carrying current 3.00

Question

The magnetic field 36.0 cm away from a long, straight wire carrying current 3.00 A is 1670 µT.

(a) At what distance is it 167 µT?

360 cm

(b) At one instant, the two conductors in a long household extension cord carry equal 3.00-A currents in opposite directions. The two wires are 3.00 mm apart. Find the magnetic field 36.0 cm away from the middle of the straight cord, in the plane of the two wires.
nT

(c) At what distance is it one-tenth as large?
cm

(d) The center wire in a coaxial cable carries current 3.00 A in one direction, and the sheath around it carries current 3.00 A in the opposite direction. What magnetic field does the cable create at points outside the cables?

0 nT

Explanation / Answer

given

d = 0.36 m

long straight wire, carryig current, I = 3 A

Bi = 1670*10^-6 T

a. for B' = 167*10^-6 T

B = 2kI/d

hence

Bi = 2kI/d

B' = 2kI/d'

Bi/B' = d'/d

d' = d(Bi/B') = 3.6 m

b. i = 3 A

opposite directions

d = 3 mm = 3*10^-3 m

at, y = 0.36 m

B = 2ki(1/(y - d/2) - 1/(y + d/2))

B = 2*10^-7*3(1/(0.36 - 1.5*10^-3) - 1/(0.36 + 1.5*10^-3)) = 0.1388913*10^-7 T = 13.998 nT

c. at distance d'

1.3998*10^-9 = 2k*3(1/(y' - d/2) - 1/(y' + d/2))

0.0023316 = d/(y'^2 - d^2/4)

1.1343158060 m = d

d. i = 3 A

at points outside cable,from ampere's circuital law

net magnetic field = 0 T

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