. Magnetic field produced by a current-carrying wire (Text Ch. 19, Ex. 57) A lon

Question

. Magnetic field produced by a current-carrying wire (Text Ch. 19, Ex. 57) A long wire is placed 2.0 cm directly below a rigidly mounted second wire (figure at right). (a) Use the right-hand source and force rules to determine whether the currents in the wires should be in (i) the same or (ii) the opposite direction so that the lower wire is in equilibrium. (It “floats.”) (b) If the lower wire has a linear mass density of 3 1.5 10 kg m and the wires carry the same current, what should be the current?

Explanation / Answer

here,

distance between the wires , d = 0.02 m

let the current in the wires be i

linear mass density is m

The currents must be in the same direction for the wires to attract each other. Since the free wire is below the fixed wire the net force on it must be up so the currents are in the same direction

m = 31.5 kg/m

for the equilibrium

m*g = u0 * i^2 /( 2 * pi*d)

31.510*9.8 = 1.26 * 10^-6 * i^2 /( 2 * pi * 0.02)

i = 5.5 * 10^3 A

the curent in the wires is 5.5 * 10^3 A

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.