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Q.

Consider two current carrying conductors in x-y plane, carrying constant current  I1and  I2 respectively as shown in the figure.Now consider two arbitrary elementary current  elements AB & CD of lengths  dl1  and  dl2 on the loop-1 and loop-2 respectively. Let dF12  be the force which the current carrying element  I2dl2exerts on the current carrying  element I1dl1  &F12 represent the net force on 1st loop by the 2nd loop, developed due to  interaction between current carrying loops. SimilarlydF21  be the force which the current 
carrying element I1dl1 exerts on the current carrying element I2dl2  &F21 represent the net  force on 2nd loop by the 1st loop, developed due to interaction between current carrying  loops. Choose the correct option(s)
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a

dF12+dF21=0

b

dF12+dF210

c

F12+F21=0

d

F12+F210

answer is B, C.

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Detailed Solution

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dB12= Magnetic field at any point on the elementary element  AB due to current carrying  element  I2dl2.
 dB12=μ04π(I2dl×r12r123)
 dF12=I2dl1×dB12=(μ04π)I1dl1×(I2dl2×r12r123
The force dF21 will be in the plane of loop and perpendicular to the current carrying  element I1dl1 (say along p^12 )
Similarly  dF21=μ04πI2dl2×(I1dl1×r21)r213 [say along p^21]
    p^12 is not parallel to , in general
So,  dF12+dF210
But  F12=μ0I1I24πL1 L2 dl1×(dl2×r12)r123
 F21=L2 L1 dl2×(dl1×r21)r213
Hence,  F12+F21=0

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