(a)Two insulated wires, each 2.24 m long, are taped together to form a two-wire

Question

(a)Two insulated wires, each 2.24 m long, are taped together to form a two-wire unit that is 2.24 m long. One wire carries a current of 6.83 A; the other carries a smaller current I in the opposite direction. The two-wire unit is placed at an angle of 63.6° relative to a magnetic field whose magnitude is 0.348 T. The magnitude of the net magnetic force experienced by the two-wire unit is 3.09 N. What is the current I?

(b)The x, y, and z components of a magnetic field are Bx = 0.150 T, By = 0.200 T, and Bz = 0.230 T. A 27.0-cm wire is oriented along the z axis and carries a current of 4.36 A. What is the magnitude of the magnetic force that acts on this wire?

Explanation / Answer

a] The perpendicular field component is B = 0.348 T*sin 63.6 degree = 0.31 T

F = B*i*L

so the net current is i = F/BL

= 3.09 N / (0.31 T * 2.24 m)

= 4.45 A

i = I + 6.83 A

So, I = (4.45 - 6.83) A

= (-) 2.38 A

Answer

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