A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be
- nB
- n2B
- 2nB
- 2n2B
Answer
(2) n2B
Solution
Length of wire is fixed: 2 r = n(2 r') r' = rn
One turn: B = 0 I2r
n turns: B' = 0 nI2r' = 0 n2 I2r
B' = n2B
Answer: (2)