Question

The magnetic field 41.0 cm away from a long, straight wire carrying current 4.00 A is 1950 nT.

(a) At what distance is it 195 nT?


(b) At one instant, the two conductors in a long household extension cord carry equal 4.00-A currents in opposite directions. The two wires are 3.00 mm apart. Find the magnetic field 41.0 cm away from the middle of the straight cord, in the plane of the two wires.

How far is the point of interest from each wire? nT

(c) At what distance is it one-tenth as large?
cm

(d) The center wire in a coaxial cable carries current 4.00 A in one direction, and the sheath around it carries current 4.00 A in the opposite direction. What magnetic field does the cable create at points outside the cables?
nT

Answer 1

Part A.

Magnetic field due to long current carrying wire is given by:

B = u0*I/(2*pi*d)

From above equation we can see that magnetic field is inversely proportional to the distance from wire. So

B1/B2 = (r2/r1)

r2 = r1*(B1/B2)

r2 = 41.0*(1950/195) = 410 cm

So at distance 410 cm magnetic field will become 195 nT

Part B.

Magnetic field due to current carrying wire is given by:

B = u0*i/(2*pi*r)

Now since current is in opposite direction, So net magnetic field 41.0 cm away from the middle of the cord will be

Bnet = B1 - B2

Bnet = u0*i/(2*pi*r1) - u0*i/(2*pi*r2)

Bnet = (u0*i/(2*pi))*(1/r1 - 1/r2)

i = current in wires = 4.00 Amp

Since distance between both wires is 3.00 mm = 0.3 cm, So from the middle of the cord

r1 = distance from cord center to field point due to wire 1 = 41 cm - 0.3 cm/2 = 40.85 cm = 0.4085 m

r2 = distance from cord center to field point due to wire 2 = 41 cm + 0.3 cm/2 = 41.15 cm = 0.4115 m

So,

Bnet = (4*pi*10^-7*4.00/(2*pi))*(1/0.4085 - 1/0.4115)

Bnet = 14.3*10^-9 T

Bnet = 14.3 nT

Part C.

At what distance magnetic field is one-tenth as large

Bnet = 14.3 nT (from part (b))

So one-tenth, B1net = 14.3/10 = 1.43 nT

Now Suppose at distance r, net magnetic field is 1.43 nT, then

B1net = B1 - B2

B1net = (u0*i/(2*pi))*(1/r1 - 1/r2)

r1 = r - d, and r2 = r + d

d = distance of wires from center of cord = 3.00 mm/2 = 1.5 mm = 0.0015 m

So,

B1net = (u0*i/(2*pi))*(1/(r - d) - 1/(r + d))

B1net = (u0*i/(2*pi))*(2d/(r^2 - d^2))

r^2 - d^2 = u0*i*d/(pi*B1net)

r^2 = d^2 + u0*i*d/(pi*B1net)

Using known values:

r^2 = 0.0015^2 + (4*pi*10^-7*4.00*0.0015/(pi*1.43*10^-9))

r^2 = 1.67832393

r = sqrt (1.67832393)

r = 1.2955 m

r = 129.6 cm

Part D.

According to Ampere's law, there will be no current outside the cables, So there will be no magnetic field.

Magnetic field outside the cables = 0 nT

Let me know if you've any query.