The magnetic field 37.0 cm away from a long, straight wire carrying current 3.00

Question

The magnetic field 37.0 cm away from a long, straight wire carrying current 3.00 A is 1620 mu T. At what distance is it 162 mu T? 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 37.0 cm away 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 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?

Explanation / Answer

a) 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*10^-7*I/(2*pi*r)
= 2x10^-7xI/r
= 2x3x10^-7/r
B(0.37(m)) = 6x10^-7/0.37 = 1.62x10^-6 (T)

162x10^-6 = 2x3x10^-7/r

or, r = 2x3x10^-7/162x10^-6 = 0.0037 m = 0.37 cm......................ans a)

b) If you are measuring 37 (cm) from the center, you are getting 37.15 (cm) from one and 36.85 (cm) from the other; and their B-fields are opposite in direction. So the answer is:
B(36.85) - B(37.15) = 2x3.0(1/0.3685 - 1/0.3715)x10^-7
= 1.314x10^-8 T

c) B(r - 1.5 mm) - B(r + 1.5 mm) = 1.314x10^-8/10 = 1.314x10^-9
6x10^-7*(1/(r - 1.5 mm) - 1/(r + 1.5 mm)) = 1.314x10^-9
(1/(r - 1.5 mm) - 1/(r + 1.5 mm)) = (1.314/6)x10^-2 = 2.19x-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:
3x10^-3/r^2 = 2.19x10^-3
=> r^2 = 3x10^-3/(2.19x10^-3) = (3/2.19) = 1.37
r = sqrt(1.37) = 1.17 (m)

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