A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will be

nB

n²B

2nB

2n²B

Solution

n²B

The magnetic field at the centre of circular coil is
$straight B space equals fraction numerator straight mu subscript 0 space straight i over denominator 2 straight r end fraction where space straight r space equals space radius space of space circle space space equals space fraction numerator 1 over denominator 2 straight pi end fraction space left parenthesis because space straight I space equals 2 πr right parenthesis straight B space equals space fraction numerator straight mu subscript 0 space straight i space over denominator 2 end fraction space straight x space fraction numerator 2 straight pi over denominator straight I end fraction space equals space fraction numerator straight mu subscript 0 space straight i space straight pi over denominator straight I end fraction space.... space left parenthesis straight i right parenthesis$
When wire of length i bents into a circular loops of n turns, then
l = n × 2 π r
⇒ r = 1/ n x 2 π
Thus, new magnetic field
$straight B apostrophe space equals space fraction numerator straight mu subscript 0 ni over denominator 2 straight r apostrophe end fraction space equals space fraction numerator straight mu subscript 0 ni over denominator 2 end fraction space straight x space fraction numerator straight n space straight x space 2 straight pi over denominator straight l end fraction space equals space fraction numerator straight mu subscript 0 space straight i space straight pi over denominator straight l end fraction space straight x space straight n squared space equals space straight n squared straight B$