A long horizontial wire carries a current of 25 A flowing to the right. A second

Question

A long horizontial wire carries a current of 25 A flowing to the right. A second horizontal wire, carrying an unknown current I, lies a distance .24m directly below the first wire.

A. What is the direction of the magnetic field produced by the top wire where the bottom wire is?

B. The Bottom wire is supported by the force of attraction due to the top wire. What is the direction of the current I in the bottom wire?

C. The bottom wire has a length of 1.4 m and is made of material whose linear desinty is .02 kg/m . what is the mass of the bottom wire?

D. The weight of the bottom wire is euql to the force exerted on it due to the top wire. By equating these tow force, find the magnitude of the current I in the bottom wire.

Explanation / Answer

A) B = mue*I/(2*pi*d)

= 4*pi*10^-7*25/(2*pi*0.24)

= 2.08*10^-5 T

B) towards right

C) m = 1.4*0.02 = 0.028 kg

d)

FB = Fg

B*I*L = m*g

I = m*g/(B*L)

= 0.028*9.8/(2.08*10^-5*1.4)

= 9423 A

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