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Questions  

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

a
x= I1I2a
b
x= I2I1a
c
x= I1−I2I1 + I2a
d
x= I1+I2I1 − I2a

detailed solution

Correct option is C

μ0I14π a+x  =  μ0I24π a−xa−xa+x = I2I1I1a − I1x = I2a+I2xx= I1−I2I1 + I2a

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