The correct option is A μ0I8πa(−^i+^k)
One part of the wire is lying along positive x-axis, one part along the line x=z
and one part along positive y-axis.
There is no magnetic field at the origin due to the segments 1 and 2 (since they both pass through the origin). Hence, net magnetic field at origin is only due to 3.
The magnetic field due to 3 at origin is given by the formula for the magnetic field due to a semi infinite wire at a point. B=μ0i4πd(sinα+sinβ)
are the angles made by the line joining the point O with the extremes of the wire.
Here, α=0∘,β=90∘ ∴B=μ0i4πd
To find the direction of field, plot the magnetic field lines around the wire passing through point O.
Then, we can say that the magnetic field at O must be perpendicular to the radius of the circle passing through O shown in the diagram.
Since the radius is along the line x=z
, the line perpendicular to it is x=−z
i.e the direction of magnetic field at O is along (−^i+^k
Thus, magnetic field at O can be written as →B=μ0i4πdcos45∘(−^i)+μ0i4πdsin45∘(^k) =μ0i4π√2acos45∘(−^i+^k) =μ0i8πa(−^i+^k)