Two circular coils $P$ and $Q$ of 100 turns each have the same radius of $\pi\ \text{cm}$. The currents in $P$ and $Q$ are $1\ \text{A}$ and $2\ \text{A}$ respectively. $P$ and $Q$ are placed with their planes mutually perpendicular with their centres coinciding. The resultant magnetic field induction at the centre of the coils is $\sqrt x\ \text{mT}$, where $x = $ ______. [Use $\mu_0 = 4\pi\times10^{-7}\ \text{T m A}^{-1}$]
Numerical value type. Enter your answer.
Answer: 20
$B = \dfrac{\mu_0NI}{2R}$ with $R = 0.01\pi\ \text{m}$:
$$B_P = \frac{4\pi\times10^{-7}\times100\times1}{2\times0.01\pi} = 2\ \text{mT},\qquad B_Q = 4\ \text{mT}$$
The fields are perpendicular: $B = \sqrt{4 + 16} = \sqrt{20}\ \text{mT}$, so $x = 20$.
Solution by Sreeraj P, M.Sc Physics