The magnetic field 43.0 cm away from a long, straight wire carrying current 8.00

Question

The magnetic field 43.0 cm away from a long, straight wire carrying current 8.00 A is 3720 µT.

(a) At what distance is it 372 µT?
_____
cm

(b) At one instant, the two conductors in a long household extension cord carry equal 8.00-A currents in opposite directions. The two wires are 3.00 mm apart. Find the magnetic field 43.0 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 8.00 A in one direction, and the sheath around it carries current 8.00 A in the opposite direction. What magnetic field does the cable create at points outside the cables?
______ nT

Explanation / Answer

a)   Magnetic field is inversely proportional to distance
So d2 = d1*B1/B2 = 0.43*3720E-6/372E-6 = 4.3 m

b) According to Ampere's law:
2*pi*r*B(r) = I*µ0 where µ0 = 4*pi*e-7
So from just one current, B(r) = 4*pi*e-7*I/(2*pi*r) = 2e-7*I/r = 2*8*e-7/r

If you are measuring 43 (cm) from the center, you are getting 43.15 (cm) from one and 42.85 (cm) from the other; and their B-fields are opposite in direction. So the answer is:
B(42.85) - B(43.15) = 16*(1/0.4285 - 1/0.4315)*e-7 = 25.96 nT

c) So we want the solution to the equation:
B(r - 1.5 mm) - B(r + 1.5 mm) = 25.96*e-9/10 = 25.96e-10
16e-7*(1/(r - 1.5 mm) - 1/(r + 1.5 mm)) = 25.96e-10
(1/(r - 1.5 mm) - 1/(r + 1.5 mm)) = (25.96/16)*e-3 = 1.62e-3

The easy way to do this is to notice that, if r is >> 1.5 mm, we can approximate the left-hand side by:
- 2 * 1.5 (mm) *d(1/r)/dr = 2*1.5 (mm)/r^2 = 3e-3/r^2 , so:
3e-3/r^2 = 1.62e-3
=> r^2 = 3e-3/(1.62e-3) =1.85
r = sqrt(1.85) = 1.36 m
Indeed, 1.36 (m) is >> 1.5 (mm), so our approximation is justified.
So the answer is: r = 1.36 (m)

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