Question

Two similar coils of radius $8cm$ and number of coils $=100$ are parallel placed. The distance between their centres is $12cm$. If $1A$ current flows in the same direction in both the coils then calculate the value of magnetic field on the axis right in the middle.

Answer 1

Radius of each coil $R=8cm=8\times {10}^{-2}m$
Number of turns $N=100$
Magnetic field at mid point the line joining two
$B_1=B_2=B$
$B=\frac{{\mu }_{0}NI{R}^2}{2.{(R^2+x^2)}^{3/2}}$
$B=\frac{{\mu }_{0}NI{R}^2}{2.{(R^2+x^2)}^{3/2}}$
$\frac{4\pi \times {10}^{-7}\times 10\times 1.0}{2\times 0.10}=\frac{4\pi \times {10}^{-7}\times 100\times 1\times {(8\times {10}^{-2})}^2}{2{[{(8\times {10}^{-2})}^2+{(6\times {10}^{-2})}^2]}^{3/2}}=\frac{4\times 3.14\times 64\times {10}^{-9}}{2\times {10}^{-3}}=4.02\times {10}^{-4}T$
$B_1$ and $B_2$ are in same direction
$\therefore$ Resultant magnetic field
$B=B_1+B_2=2B$
$B=2\times 4.02\times {10}^{-4}$
$=8.04\times {10}^{-4}T$