A straight wire carrying a current of 12 A is bent into a semicircular arc of radius 2.0 cm as shown in figure (a). Consider the magnetic field `vecB` at the centre of the arc.
(a) What is the magnetic field due to the straight 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 semicircular arc of the same radius but in the opposite way as shown in figure (b) ?

Explanation

(a)
In both of above figures, point of observation (here centre P of circular arc where we want to find out magnetic field) is colinear with straight current carrying parts ab and de. Hence, for these parts,
`|vec(dl)xxvecr|=(dl)(r)sintheta=0" "(becausetheta=0^(@)or180^(@))`
`rArrvec(dl)xxvecr=vec0`
Hence, these parts do not produce any magnetic field at point P.
(b) Here, all the current elements. chosen on semicircular arc bcd, produce magnetic field at point P in the same direction (perpendicularly inside the plane of figure) and so we can make scalar addition of partial magnetic fields produced by them. Value of this addition is half of that produced by one turn of a circular loop. Hence, magnetic field produced by semi circular arc bcd at centre point P will be,
`B=1/2((mu_(0)I)/(2R))`
`=((4pixx10^(-7))(12))/((4)(0.02))`
`thereforeB=1.884xx10^(-4)T`
(Direction : perpendicularly inside the plane of figure).

( c) For semicircular are given in figure (b), magnetic field produced at centre P has same magnitude `B=1.884xx10^(-4)T` but its direction is just opposite to previous magnetic field. Thus in this case, `B=1.884xx10^(-4)T`
(Direction : perpendicularly outside the plane of figure.)