A straight wire carrying a current of `12A` is bent into a semicircular arc of radius `2*0cm` as shown in figure. Consider the magnetic field `vecB` at the centre of arc.
(a) What is the magnetic field due to the staight segments?
(b) In what way the contribution to `vecB` from the semicircle differs from that of a circular loop and in what way does it resemble?
(c) Would your answer be different if the wire were bent into a semicircle arc of the same radius but in the opposite way as shown in figure

Explanation

(a) For a point O, `dvecl` and `vecr` for each element of the straight segments AB and DE are parallel. Therefore, `dveclxxvecr=0`. Hence, magnetic field at O due to current through straight segments AB and DE is
`vecB=sum(mu_0)/(4pi)(Idveclxxvecr)/(r^3)=0`
(b) The magnetic field induction at O due to semicircular arc is
`B=(mu_0)/(4pi)(Ipi)/(r)=(mu_0)/(4)I/r=(4pixx10^-7xx12)/(4xx2xx10^-2)`
`=1*9xx10^-4T`
The direction of `vecB`, according to Right hand rule is normal to the plane of paper directed inwards. The magnetic field induction at the centre of a circular loop of radius r carrying current I is,
`B^'(mu_0)/(4pi)(2piI)/(r)=(mu_0I)/(2r)=(4pixx10^-7xx12)/(2xx2xx10^-2)`
`=3*8xx10^-4T`
It is double than that due to semi-circular arc. The direction of magnetic field induction at the centre of circular arc is the same as that due to semicircular arc if current in them is same and in same direction (i.e., either clockwise or anticlockwise) and the direction of magnetic field induction at the centre of circular arc is different from that due to semicircular arc if current in them is in opposite directions.
(c) When the wire is in the form shown in figure, the magnetic field induction at O will remain same in magnitude `(=1*9xx10^-4T)` but its direction will be normal to the plane of paper directed outwards.