The magnetic field at point P due to a 2.0-A current flowing in a long, straight

Question

The magnetic field at point P due to a 2.0-A current flowing in a long, straight, thin wire is 8.0 mu T. How far is point P from the wire? (mu 0 = 4 pi Times 10-7 T middot m/A) 2.0 cm 2.5 cm 4.0 cm 5.0 cm 10 cm Two long parallel wires that are 0.40 m apart carry currents of 10 A in opposite directions. What is the magnetic field strength in the plane of the wires at a point that is 20 cm from one wire and 60 cm from the other? (mu 0 = 4 pi Times 10-7 T middot m/A) 3.3 mu T 6.7 mu T 33 mu T 67 mu T Two long, parallel wires carry currents of different magnitudes. If the current in one of the wires is doubled and the current in the other wire is halved, what happens to the magnitude of the magnetic force that each wire exerts on the other? It is doubled. It stays the same. It is tripled. It is quadrupled. It is reduced by a factor of two.

Explanation / Answer

28) B = mu_0*i/(2*pi*r) = 8*10^-6

mu_o/2*pi = 2*10^-7 T.m/A

then (2*10^-7*2)/r = 8*10^-6

r = 0.05 m = 5 cm

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29) B1 = mu_0*i1/(2*pi*r)=(2*10^-7*10)/(0.2) = 1*10^-5 T = 10*10^-6 T

B2 = mu_o*i2/(2*pi*r) = (2*10^-7*10)/(0.6) = 3.33*10^-6 T

then Bnet = B1-B2 = (10-3.3)*10^-6 = 6.7*10^-6 T = 6.7 uT

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force F = mu_o*i1*i2*l/(2*pi*r)

now i1'= 2*i1

i2' = i2/2

then New force F' = mu_o*i1'*i2'*l/(2*pi*r)

F' = mu_o*(2*i1)*(i2/2)*l/(2*pi*r)

F' = mu_o*i1*i2*l/(2*pi*r) = F

B) It stays the same.

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