Question

A current of $0.24A$ flows through a circular coil of $72$ turns, the average diameter of the coil being $20$ cm. What is the strength of field produced at the centre of the coil?

• A. $1.09\times {10}^{-4}T$
• B. $1.09\times {10}^{-3}T$
• C. $1.09\times {10}^{-2}T$
• D. $1.5\times {10}^{-4}T$

Answer 1

The correct option is A $1.09\times {10}^{-4}T$

$I=0.24A$
$n=72turns$
$d=20cm;r=10cm$
Magnetic field due to circular loop
$B=\frac{{\mu }_{0}ni{r}^2}{2(r^2+x^2{)}^{\frac{3}{2}}}$ at centre $x=0$
$\therefore B=\frac{{\mu }_{0}ni}{2r}=\frac{4\pi \times {10}^{-7}\times 72\times 0.24}{2\times \left(0.1\right)}$
$=1.09\times {10}^{-4}T$