A thin copper bar of length l = 14.0 cm is supported horizontally by two (nonmag

Question

A thin copper bar of length l = 14.0 cm is supported horizontally by two (nonmagnetic) contacts at its ends. The bar carries a current of I1 = 125 A in the negative x direction, as shown in the figure below. At a distance h = 0.500 cm below one end of the bar, a long, straight wire carries a current I2 = 200 A in the positive z direction. Determine the the magnetic force exerted on the bar. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.

Explanation / Answer

the magnetic force on the bar of length dx is,

dF=i1*B*dx*sin(theta)

here, magnetic field B at r due to a long thin wire is,

B=(uo*i2/2pi*r)

now,

dF=i1*dx*B*sin(theta)

dF=(uo*i1*i2/2pi*r)*x*dx/r

dF=(uo*i1*i2/2pi)*(xdx/r^2)

the net force on the bar is,

F=integration of dF (between the limit x=0 to x=l)

F=(uo*i1*i2/2pi)*itegral of [x*dx/x^2+h^2)] (between the limit x=0 to x=l)

F=(uo*i1*i2/2pi)ln[(h^2+l^2)/h^2]

= (4pi*10^-7*125*200/2pi)ln[0.5^2+14^2/0.5^2)

= (4pi*10^-7*125*200/2pi)ln[196.25/0.25]

F = 0.033 N

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