A wire with a weight per unit length of 0.050 N/ms is suspended directly above a

Question

A wire with a weight per unit length of 0.050 N/ms is suspended directly above a second wire. The top carries a current of 18.0 A and the bottom wire carries a current of 23.0 a. Find the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion. A. 8.28 mm B. 1.66 mm C. 0.121 mm D. 12.9 mm E. 311 mm A loop of wire moving to the right passes through a region of space where a magnetic field is present. At which numbered position is the induced current in the loop clockwise? A. Position 2 B. Position 3 C. Position 4 D. Positions 2 and 4 E. Positions 2, 3, and 4 A conducting rod of length 0.30 m and resistance 10.0 Ohm moves with a speed of 2.0 m/s through a magnetic field of 0.20 T which is directed out of the page. The current produced in the rod is A. 0.012 A B. 0.040 A C. 0.60 A D. 0.072 a E. 0.40 A

Explanation / Answer

magnetic force per nit length between wires Fb = uo*I1*I2/(2*pi*r)

I1 = 23 A

I2 = 18 A

r = seperation

magnetic force Fb = weight per unit length

uo*I1*I2/(2*pi*r) = 0.05

4*pi*10^-7*23*18/(2*pi*r) = 0.05

r = 1.66 mm <<<--------answer

option (B)

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11)

position 4

OPTION ( C) <<<------answer

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12)

induced emf = e = B*v*l

current i = e/R

i = B*v*l/R

i = 0.2*2*0.3/10

i = 0.012 A <<<<---------answer

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