Two infinitely long straight wires lie in the xy-plane along the lines `x =pm R`. The wire located at `x =+R` carries a constant current `I_(1)` and the wire located at ` x = -R` carries a constant current `I_(2)`. A circular loop of radius R is suspended with its centre at `(0,0, sqrt(3)R)` and in a plane parallel to the xy-plane. This loop carries a constant current ? in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the `+hat J` direction. Which of the following statements regarding the magnetic field ` vec B` is (are) true?

Options

A. If `I_(1)=I_(2)`, then `vecB`cannot be equal to zero at the origin (0, 0, 0)
B. If `I_(1)gt0` and `I_(2)lt0`, then `vecB` can be equal to zero at the origin (0, 0, 0)
C. If `I_(1)lt0` and `I_(2)gt0` then `vecB`can be equal to zero at the origin (0, 0, 0)
D. If `I_(1)=I_(2)`, then the z-component of the magnetic field at the centre of the loop is `(-(mu_(0)I)/(2R))`

Answer

A, B, D

Explanation

B due to `I_(1)` at origin `=(mu_(0)I_(1))/(2piR)(+hatk)`, B due to `I_(2)` at origin `=(mu_(0)I_(2))/(2piR)(-hatk)`
B due to I at origin `=(mu_(0)I)/(16R)(-hatk)`
`barB_("(origin")=(mu_(0))/R((I_(1))/(2pi)-(I_(2))/(2pi)-(I/16))hatk`
So if `i_(1)=I_(2)impliesB!=0implies` A is correct
So if `I_(1)gtI_(2)implies` I will have a value for which B can be zero `implies` B is correct
so if `I_(1)lt0,I_(2)gt0impliesB_("origin")` has to be alogn `-hatkimplies` C is incorrect
If `I_(1)=I_(2)implies` B at the centre of loop `=(-(mu_(0)I)/(2R))hatkimplies` D is correct.
A Z- component of B due to `I_(1)` & `I_(2)` cancel out.