The magnetic field 35.0 cm away from a long, straight wire carrying current 1.80
ID: 1541137 • Letter: T
Question
The magnetic field 35.0 cm away from a long, straight wire carrying current 1.80 A is 1.03 T.
(a) At what distance is it 0.103 T?__________ cm
(b) At one instant, the two conductors in a long household extension cord carry equal 1.80 A currents in opposite directions. The two wires are 3.00 mm apart. Find the magnetic field 35.0 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 1.80 A in one direction, and the sheath around it carries current 1.80 A in the opposite direction. What magnetic field does the cable create at points outside?___________ nT
Explanation / Answer
a) B = uI / 2*pi*d
1.03 uT at 35 cm
0.103 uT at 3.5 m
b) If you are measuring 35 (cm) from the center, you are getting 35.15 (cm) from one and 34.85 (cm) from the other; and their B-fields are opposite in direction. So the answer is:
B(34.85) - B(35.15) = 3.6*(1/0.3485 - 1/0.3515)*e-7
= 8.82*e-9 (T) = 8.82 nT
c) So we want the solution to the equation:
B(r - 1.5 mm) - B(r + 1.5 mm) = 8.82*e-9/10 = 8.82e-10
3.6e-7*(1/(r - 1.5 mm) - 1/(r + 1.5 mm)) = 8.82e-10
(1/(r - 1.5 mm) - 1/(r + 1.5 mm)) = (8.82/3.6)*e-3 = 2.45e-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 = 2.45e-3
=> r^2 = 3e-3/(2.45e-3) = (3/2.45) = 1.2245
r = sqrt(1.2245) = 1.11 (m)
d) 0
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.