A closely packed coil having 1000 turns has an average radius of 62.8 cm. If the current carried by the wire of the coil is 1 A, the value of magnetic field produced at the centre of the coil will be (permeability of free space $=4\pi\times10^{-7}$ H/m) nearly:

A $10^{-3}$ T

B $10^{-1}$ T

C $10^{-2}$ T

D $10^{2}$ T

The answer is A.

Solution

$B=\frac{\mu_0 NI}{2R}=\frac{4\pi\times10^{-7}\times1000\times1}{2\times62.8\times10^{-2}}=10^{-3}$ T.

Concept Magnetic field at the centre of a circular coil

15 questions from this chapter have appeared across 12 exam years.