Questions 10 - 13 please. Two very long wires both carry current toward the west
ID: 1499924 • Letter: Q
Question
Questions 10 - 13 please.
Two very long wires both carry current toward the west One is directly above the other, and they are 2.0 cm apart. The top wire (hanging by a series of insulating support cables) carries a current UP of 300 A. The bottom, resting on top of a long insulating table, carries a current of 100 A. What is the magnetic field at a point 2.0 cm north of the top wire due to: the top wire only? 3.0 mT[D] the bottom wire only? 0.707 mT[N45degree D] both wires? 3.54 mT[N82 degree D] What is the total magnetic field at a point 2.0 cm north of the bottom wire? 2.9 mT [D.31 degree S] What is the magnetic field halfway between the wires? 4.0 mT [south] What is the magnetic field sensed by the bottom wire? 3.0 mT [south] What magnetic force (per unit length) acts on the bottom wire? 0.30 N/m [up] If there is no normal force acting on the bottom wire, what must its mass (per unit length) be? 0.0306 kg/m What is the magnetic field sensed by the top wire? 1.0 mT [north] What magnetic force (per unit length) acts on the top wire? Does your answer agree with Newton's Third Law? 0.30 N/m [down]Explanation / Answer
10)
Itop = 300 A
Ibottom = 100 A
(a)
Magnetic Field due to current carrying wire is given by, B = (uo*I)/(2**r)
B = (4**10^-7 * 300) / (2**2.0*10^-2)
B = 0.003 T
B = 3.0 mT @ D
(b)
r = sqrt(2.0^2 + 2.0^2)
r = 2.83 * 10^-2 m
Due to bottom wire,
B = (4**10^-7 * 100) / (2**2.83*10^-2)
B = 0.000707 T
B = 0.707 mT @ N45o D
(c)
Due to both the wire,
B1 = - 3.0 j^ mT
B2 = 0.707 * cos(45) + 0.707 * sin(45) mT
B2 = - 0.5j^ + 0.5 i^ mT
Bnet = sqrt( (0.5) + (0.5 + 3.0)^2 )
Bnet = 3.54 mT
Angle = tan^-1(3.5/0.5) = 81.87o
Bnet = 3.54 mT @ N 82.0o D
11)
Due to bottom wire,
B1 = (4**10^-7 * 100) / (2**2.0*10^-2)
B1 = 0.001 T
B1 = 1 mT
Due to top wire,
B2 = (4**10^-7 * 300) / (2**2.83*10^-2)
B2 = 0.00212 T
B2 = 2.12 mT
Bnet = sqrt( (2.12*cos(45) )^2 ) + ( (2.12*cos(45) + 1.0 )^2 ) )
Bnet = 2.92 mT
Angle = tan^-1((2.12*cos(45) + 1.0 )/ (2.12*cos(45) ))
Angle = 90 - 59 = 31o
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