Consider a tightly wound 100 turn coil of radius $10 \mathrm{~cm}$ carrying a current of $2 \mathrm{~A}$.…
Consider a tightly wound 100 turn coil of radius $10 \mathrm{~cm}$ carrying a current of $2 \mathrm{~A}$. The magnitude of the magnetic field at the centre of the coil is
$3.14 \times 10^{-4} \mathrm{~T}$
$6.28 \times 10^{-4} \mathrm{~T}$
$12.56 \times 10^{-4} \mathrm{~T}$
$0$
Solution
Number of turns $n=100$
Radius, $r=10 \mathrm{~cm}=0.1 \mathrm{~m}$
current, $\mathrm{I}=2 \mathrm{~A}$
Magnetic field at center of coil:
$\begin{aligned} & \mathrm{B}=\frac{\mu_0 \mathrm{ni}}{2 \mathrm{r}}=\frac{4 \pi \times 100 \times 2 \times 10}{2 \times 0.1} \times 10^{-7} \\ & \mathrm{~B}=12.56 \times 10^{-4} \mathrm{~T}\end{aligned}$