Question

$A$ circular coil of radius $10\, cm$ having $100$ turns carries a current of $3.2 \,A$. The magnetic field at the center of the coil is

• A. $2.01 \times 10^{-3}\, T$
• B. $5.64 \times 10^{-3}\, T$
• C. $2.64 \times 10^{-4}\, T$
• D. $5.64 \times 10^{-4} \, T$

Answer 1

Answer: (A)

As $B=\frac{\mu_{0}\,NI}{2R}$,
Here $N = 100$,
$I=3.2 A$,
$R= 10 cm = 10\times10^{-2}\,m$
$\therefore B=\frac{4\pi\times10^{-7}\times100\times3.2}{2\times0.1}$
$=2.01\times10^{-3}\,T$