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 magnetic field at the centre of the coil is

• A. $2.4 \times 10^{-4}\, T$
• B. $3.5 \times 10^{-4}\, T$
• C. $2.79 \times 10^{-4}\, T$
• D. $3 \times 10^{-4}\, T$

Answer 1

Answer: (C)

Here $N= 100$
$R=9\, cm =9\times10^{-2}\,m$ and $ I=0.4\,A $
Now, $B=\frac{\mu_{0}NI}{2R}=\frac{2\pi\times10^{-7}\times100\times0.4}{9\times10^{-2}}$
$=\frac{2\times3.14\times0.4}{9}\times10^{-3} =0.279\times10^{-3} $
$T=2.79\times10^{-4}\, T$