A rectangular loop of wire of dimension a?b faces a long, straight wire, as show

Question

A rectangular loop of wire of dimension a?b faces a long, straight wire, as shown on the figure below. The two long sides of the loop are parallel to the wire, and the two short sides are perpendicular; the midpoint of the loop is distance d away from the wire. Currents of I1 and I2 flow in the straight wire and the loop, respectively. (a) What translational force does the straight wire exert on the loop? (b) What torque about an axis parallel to the straight wire and through the center of the loop does the straight wire exert on the loop?

Explanation / Answer

The field is parallel to the shorter sides, so force on tghe short sides is 0

Force on the long side = i (L x B)
consider the upper long side
|B| = 2kI/r [ magnetic field due to a long wire]
Now r = sqroot(d^2 + (b/2)^2).. (1)

and |F| = I2 * a * 2k I1 / r .. (2)
direction of force, alng the line joining the wires, repulsive in nature
let intop the plane be x - direction ( i ) and in the plane (positive upwards) be y - direction ( j )
so angle between x - direction and force vector is, theta = arctan(b/2d)
tan(theta) = b/2d
sin(theta) = b/2r
cos(theta) = d/r

so force, F = |F|[icos(theta) + jsin(theta)] = |F|[2di + bj]/2r = 2k*I1*I2*a[2di + bj]/2r^2

similiarly force on the lower long wire = 2k*I1*I2*a[-2di + bj]/2r^2

a)Net translational force = F1 + F2 = 2k*I1*I2*a[2di + bj]/2r^2 + 2k*I1*I2*a[-2di + bj]/2r^2 = 2k*I1*I2*a[bj]/r^2 = 2k*I1*I2*a[bj]/(d^2 + (b/2)^2)
b)Net torque = force along x axis * arm of the coil = 2k*I1*I2*a[di]*b/r^2 = 2k*I1*I2*a*b[di]/(d^2 + (b/2)^2)

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