The magnetic field 42.0 cm away from a long, straight wire carrying current 4.00

Question

The magnetic field 42.0 cm away from a long, straight wire carrying current 4.00 A is 1900 mu T. At what distance is it 190 mu T? 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 42.0 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? At what distance is it one-tenth as large? You will need to do a little algebra to get this result. Try to work out a general equation for the magnetic field as a function of r and d (the separation between the wires), cm 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?

Explanation / Answer

masgnetic field B = uo I /(2pi R)

part B:

Bnet = B1 + B2

B1 is at distance of 42cm + 0.15 cm = 42.15 cm

B2 is at distance of 42 + 0.3 = 42.3 cm

so

Bnet = (4pi*10^-7/(2pi)*(4/0.4215+ 4/0.423)

Bnet = 3789 nT

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1/10 of 1900 uT = 190 uT

B = 190 *10^-6 = ( 4pi *10^-7* 4)/(2pi r)

r = 0.42 cm

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