The magnetic field 41.0 cm away from a long, straight wire carrying current 9.00

Question

The magnetic field 41.0 cm away from a long, straight wire carrying current 9.00 A is 4390 µT.

(a) At what distance is it 439 µT?
1   cm

(b) At one instant, the two conductors in a long household extension cord carry equal 9.00-A currents in opposite directions. The two wires are 3.00 mm apart. Find the magnetic field 41.0 cm away from the middle of the straight cord, in the plane of the two wires.
232.12   nT

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

(d) The center wire in a coaxial cable carries current 9.00 A in one direction, and the sheath around it carries current 9.00 A in the opposite direction. What magnetic field does the cable create at points outside the cables?

Explanation / Answer

a)

B =uoI/2pir

Given that

B =439uT =439*10-6T

I =9A

uo =4pi*10-7H/m

Now B =uoI/2pir ===>439*10-6T = 2*10-7*9A/r

Then the distance (r) =2*10-7*9A/439*10-6T =4.10mm =0.41cm

b)

The current passing through the two conductors is i =9A

The sepearation between the wire is (r) =3mm=3*10-3m =0.3cm

The magnetic field at a distance of 41cm is given by

The total magnetic field is givenby

B =kI2d/(L2-d2) =2*10-7(9)(0.003)/(0.412 -0.00152) =0.321*10-7T=32.1nT

d)

The answer iszero.This is because the currents are symmetrical, i.e. coaxial, sothey have the same B field at all distances in opposite directions.The total is zero everywhere.

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