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

x=(I1I2)a

x=(I2I1)a

x=(I1−I2I1+I2)a

x=(I1+I2I1−I2)a

A

x=(I1−I2I1+I2)a

B

x=(I1+I2I1−I2)a

C

x=(I1I2)a

D

x=(I2I1)a

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Solution

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Magnetic field with both wire will be equal and opposite in diretion,

μ0I14π(a+x)=μ0I24π(a−x)

a−xa+x=I2I1

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