What is the net force on the rectangular loop of wire that is 2 cm wide, is 6 cm

Question

What is the net force on the rectangular loop of wire that is 2 cm wide, is 6 cm long, and is located 2 cm from a long straight wire that carries 40 A of current as shown in the figure below? Assume a current of 20 A is in the loop. (Let the +x axis point to the right, the +y axis point up toward the top of the page and the +z axis point out of the page toward you.) A long, straight wire that carries a current i_wire = 40 A is to the left of a square loop of wire of length l = 0.060 m that is carrying a current of i_loop = 20 A. The left side of the loop is a distance r_left = 0.020 m from the wire; the right side of the loop is a distance r_right = 0.040 m from the wire. The net force due to the long, straight wire on the loop is equal to the vector sum of the forces acting on each segment of the loop due to the magnetic field created by the straight wire. The force on each segment of the wire can be calculated using F = A Times B and the right hand rule, where i is the current in the loop and B is the magnetic field due to the long wire at the location of the loop segment.

Explanation / Answer

Net force on recatngular loop,

Fnet = mue*I1*I2*L/(2*pi*d1) - mue*I1*I2*L/(2*pi*d2)

= mue*I1*I2*L/(2*pi)*(1/d1 - 1/d2)

= 4*pi*10^-7*20*40*0.06/(2*pi)*(1/0.02 - 1/0.04)

= 2.4*10^-4 N (attractive force)

so, in vector notation

Fnet = -2.4*10^-4 j + 0 j + 0 k

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