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Q.

Currents I1, and I2, flow in the wires shown in figure. The field is zero at distance x to the right of O. Then

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a

x=I1I2a

b

x=I2I1a

c

x=I1-I2I1+I2a

d

x=I1+I2I1-I2a

answer is C.

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Detailed Solution

μ0I14π(a+x)=μ0I24π(a-x) a-xa+x=I2I1⇒x=I1-I2I1+I2a
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