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

Step-by-step solution

Suppose the length of wire is l Then, for the circle of n turns The magnetic field, B = μ 0 I n 2 R , where R is the radius of the circle Then B s = μ 0 I 2 R =B......(i) Here, 2 π R = l .......(ii) Now, if the same wire is turned into a coil of n turns of radius a then, 2 π a × n = l .......(iii) Thus, 2 π R = 2 π a × n ⇒ a = R n Now, the magnetic field for the coil of n turns, is B n = μ 0 I 2 a × n = μ 0 I 2 R × n 2 ∵ a = R n = n 2 B