Question

A circular coil of wire consisting of $100$ turns each of radius $9 \,cm$ carries a current of $0.4 A$. The magnitude of the magnetic field at the centre of coil is [$\mu_{0} =12.56 \times 1^{-7}$ SI Unit]

• A. $2.4 \times 10^{-11} T$
• B. $2.79\times 10^{-5} T$
• C. $2.79 \times 10^{-4} T$
• D. $2.79 \times 10^{-3} T$

Answer 1

Answer: (C)

Given, number of turns, $n=100$,
radius of coil, $r=9 \,cm =9 \times 10^{-2} m$
and current flow in the coil, $I=0.4 A$
The magnitude of the magnetic field at the centre of a coil of $n$ turns is given by
$B=\frac{\mu_{0} n I}{2 r}\,\,\,\,\,\,...(i)$
Substituting given values in Eq. (i), we get
$\Rightarrow \,\,\,\ B =\frac{1256 \times 10^{-7} \times 100 \times 0.4}{2 \times 10^{-1} \times 9 \times 10^{-2}} $
$=2.79 \times 10^{-4} T$