The magnetic field 43.0 cm away from a long, straight wire carrying current 1.00

Question

The magnetic field 43.0 cm away from a long, straight wire carrying current 1.00 A is 470 µT.

(a) At what distance is it 47 µT?
430 cm

(b) At one instant, the two conductors in a long household extension cord carry equal 1.00-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 1.00 A in one direction, and the sheath around it carries current 1.00 A in the opposite direction. What magnetic field does the cable create at points outside the cables?
nT

Explanation / Answer

(a)

Magnetic field due to a long straight wire carrying current I at a distance d is;

B =  µ0I/2d

let the field be B = 47X10-6T at a distance d from the wire. So,

47X10-6T = (1.256X10-6T-m/A)(1 A)/2d ----------- (1)

at adistance 43 cm the field is 470X10-6T, so

470X10-6T = (1.256X10-6T-m/A)(1 A)/2(43 cm) ------------ (2)

or 10X(1.256X10-6T-m/A)(1 A)/2d = (1.256X10-6T-m/A)(1 A)/2(43 cm)

or d = 430 cm

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(b)

We use the equation;

B =  µ0I/2d

For one wire the distance at which the field is to be found will be d1 = 43cm +1.5mm = 43.15cm

For the other wire the distance will be d2 = 43cm - 1.5mm = 42.85 cm

So the field due to the two wires at distance 43 cm from the middle of the chord will be,

B =  µ0I/2d2 -  µ0I/2d1 = (µ0I/2)[1/d2 - 1/d1]

B = [(1.256X10-6 T-m/A)(1 A)/(6.28)][1/0.4285 m - 1/0.4315 m]

B = 3.24 nT

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(c)

Let at distance d it is 1/10th as large, so in this case,

d1 = d + 0.0015 and,

d2 = d - 0.0015m

so we have,

0.324X10-9T = (µ0I/2)[1/d2 - 1/d1] = (µ0I/2)[1/(d - 0.0015m) - 1/(d + 0.0015m)]

or, 0.324X10-9T = (µ0I/2)[2X0.0015m/(d2 - (0.0015m)2)]

or, 0.324X10-9T = [(1.256X10-6 T-m/A)(1 A)/(6.28)][2X0.0015m/(d2 - (0.0015m)2)]

or, d = 1.36 m

or d = 136 cm

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(d)

Outside the cable when we draw an Amperian loop of radius r, the current crossing the loop would be zero as the two currents inside the loop are equal and oppsite.

So the integral of B.dl on the circumference of the loop would be zero. From symmetry any field on the Amperian loop outside the cable would be circumferential (along the circumference of the loop considered) and equal in magnitude everywhere on the loop, which causes B and dl having same direction as we go around the loop and thus B.dl = Bdl, B comes out of the integral as it is constant all over the loop and we get.

B.2r = 0

or B = 0 nT

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

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