The magnetic field 44.0 cm away from a long, straight wire carrying current 1.00

Question

The magnetic field 44.0 cm away from a long, straight wire carrying current 1.00 A is 450 (a) At what distance is it 45 (b) At one instant, the two conductors in a long household carry the magnetic field 44.0 cm away from the middle of the straight cord, in the plane of the two nT extension cord carry equal 1.00-A currents in opposite directions. The two wires are 3.00 mm apart. Find (d) The center wire in a coaxial cable carries current 1.00 A in one direction, and the sheath around it field does the cable create at points outside the cables? carries current 1.00 A in the opposite direction. What magnetic

Explanation / Answer

Given,

B = 450 uT ; I = 1 A ; r = 44 cm = 0.44 m

(a)We know that

B = u0 I/2 pi R

B' = u0 I/ 2 pi R'

on diving we get

R' = r B/B'

R' = 44 x 450/45 = 440 cm

Hence, R' = 440 cm

b)In this case,

Bx = u0 I/ pi (2R + d)

By = u0 I/pi (2R - d)

Bnet = By - Bx

Bnet = u0 I/ 2pi x 2d/(2R - d)(2R + d)

Bnet = 4 pi x 10^-7 x 1/2 pi x 0.003/(2 x 0.44 - 0.003)(2 x 0.44 + 0.003) = 7.75 x 10^-10 T

Hence, Bnet = 0.775 nT

c)(2R - d)(2R + d)/(2R' - d) (2R' + d) = 1/10

Solving for R' we get

R' = sqrt (40 R^2 - 9d^2)/2))

R' = sqrt (40 x 0.44^2 - 9 x 0.003)/2)= 1.3889 m

Hence, R' = 138.9 cm

d)B = 0

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