(9%) Problem 11: Your apprentice electrician has been instructed to locate a lon

Question

(9%) Problem 11: Your apprentice electrician has been instructed to locate a long, straight, vertical, current carrying wire inside a wall. The wire carries a DC current of I-12.9 A. The electrician knows that the materials used to construct the wall do not change the magnetic permeability. In his toolbox, the electrician has an orienteering compass and a very sensitive magnetic field sensor. In this problem, recall that 1G = 10-4 T. current carrying Wire Otheexpertta.com A 20% Part (a) How far from the wire (in cm) would the magnetic field due to the current be exactly 1 Gauss? O A 20% Part (b) The blueprints specify that the current-carrying wire is supposed to be at a depth of r = 10.8 cm from the surface of the wall. Calculate the magnetic field strength (in Gauss) at a distance rf = 10.8 cm from the wire. A 20% Part (c) The electrician pulls out his orienteering compass to try to locate the wire using the magnetic field from it. As his supervisor, why would you recommend that he put his compass back in the toolbox? A 20% Part (d) The electrician uses his magnetic field sensor to locate a point on the wall with the strongest magnetic field caused by the current carrying wire. If he moves his sensor up from this spot, what will the reading on his sensor indicate? A 20% Part (e) The clectrician measures that the maximum field strength at the surface of the wall is B = 1.26 gauss. Calculate the actual depth of the wire (in cm) from the wall surface. Grade Summary Deductions Potential 100%

Explanation / Answer

(a)

Magnetic field to current carrying wire,

B = uo*l / 2*pi*r

so, distance r = uo*l / 2*pi*B

r = 4*pi*10^(-7)*12.9 / 2*pi*10^(-4) [since 1 G = 10^(-4) T]

r = 2.58 cm

(b)

Magnetic field strength,

B = uo*l / 2*pi*rb

B = 4*pi*10^(-7)*12.9 / 2*pi*(0.108)

B = 238*10^(-7) T

B = 0.238 G

(c)

He should put his compass back in toolbox. because here magnetic fiels is very small so reading of compass will be

affected by earth's magnetic field and compass will give a wrong reading.

(d)

Reading will not be affected with sensor's position (only affected by change in magnetic field). so, reading will be

same.

(e)

Actual depth of wire from the wall,

ra = uo*l / 2*pi*B

ra = 4*pi*10^(-7)*12.9 / 2*pi*1.26*10^(-4)

ra = 2.04 cm

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