Chapter 21, Problem 36 Two insulated wires, each 3.05 m long, are taped together
ID: 1259756 • Letter: C
Question
Chapter 21, Problem 36
Two insulated wires, each 3.05 m long, are taped together to form a two-wire unit that is 3.05 m long. One wire carries a current of 7.64 A; the other carries a smaller current I in the opposite direction. The two-wire unit is placed at an angle of 67.2o relative to a magnetic field whose magnitude is 0.486 T. The magnitude of the net magnetic force experienced by the two-wire unit is 2.84 N. What is the current I?
Chapter 21, Problem 36
Two insulated wires, each 3.05 m long, are taped together to form a two-wire unit that is 3.05 m long. One wire carries a current of 7.64 A; the other carries a smaller current I in the opposite direction. The two-wire unit is placed at an angle of 67.2o relative to a magnetic field whose magnitude is 0.486 T. The magnitude of the net magnetic force experienced by the two-wire unit is 2.84 N. What is the current I?
Chapter 21, Problem 36 Two insulated wires, each 3.05 m long, are taped together to form a two-wire unit that is 3.05 m long. One wire carries a current of 7.64 A; the other carries a smaller current I in the opposite direction. The two-wire unit is placed at an angle of 67.2o relative to a magnetic field whose magnitude is 0.486 T. The magnitude of the net magnetic force experienced by the two-wire unit is 2.84 N. What is the current I?Explanation / Answer
Given
magnitude of magnetic field is B = 0.486 T
angle between B and L is 67.2 degrees
length of two wire unit is 3.05 m
One wire carries a current of 7.64 A; the other carries a smaller current I in the opposite direction
The net current be
I net = ( 7.64 - I )
Force on a wire due to magnetic field is
F = I net B L sin
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