Two circular coils $$P$$ and $$Q$$ of 100 turns each have same radius of $$\pi$$ cm. The currents in $$P$$ and $$Q$$ are 1 A and 2 A respectively. $$P$$ and $$Q$$ are placed with their planes mutually perpendicular with their centers coincide. The resultant magnetic field induction at the center of the coils is $$\sqrt{x}$$ mT, where $$x =$$ [Use $$\mu_0 = 4\pi \times 10^{-7}$$ T m A$$^{-1}$$]

A circular coil of $$N$$ turns and radius $$R$$ produces, at its centre, a magnetic field along the axis of the coil given by
$$B = \frac{\mu_0 N I}{2R} \quad -(1)$$
where $$I$$ is the current through the coil.

For coil $$P$$:
Number of turns, $$N_P = 100$$
Current, $$I_P = 1 \, \text{A}$$
Radius, $$R = \pi \, \text{cm} = \pi \times 10^{-2} \, \text{m}$$

Substituting in $$(1)$$,
$$B_P = \frac{\mu_0 N_P I_P}{2R}$$
$$= \frac{(4\pi \times 10^{-7}) (100)(1)}{2(\pi \times 10^{-2})}$$
$$= \frac{4\pi \times 10^{-5}}{2\pi \times 10^{-2}}$$
$$= 2 \times 10^{-3} \, \text{T}$$
$$= 2 \, \text{mT} \quad -(2)$$

For coil $$Q$$:
Number of turns, $$N_Q = 100$$ (same as $$P$$)
Current, $$I_Q = 2 \, \text{A}$$

Again using $$(1)$$,
$$B_Q = \frac{\mu_0 N_Q I_Q}{2R}$$
$$= \frac{(4\pi \times 10^{-7}) (100)(2)}{2(\pi \times 10^{-2})}$$
$$= \frac{8\pi \times 10^{-5}}{2\pi \times 10^{-2}}$$
$$= 4 \times 10^{-3} \, \text{T}$$
$$= 4 \, \text{mT} \quad -(3)$$

The planes of the two coils are mutually perpendicular, so the fields $$B_P$$ and $$B_Q$$ are perpendicular to each other. Hence, the resultant field at the common centre is obtained by vector addition using the Pythagoras theorem:

$$B_{\text{res}} = \sqrt{B_P^{\,2} + B_Q^{\,2}}$$
$$= \sqrt{(2 \, \text{mT})^{2} + (4 \, \text{mT})^{2}}$$
$$= \sqrt{4 + 16} \, \text{mT}$$
$$= \sqrt{20} \, \text{mT}$$

The question states $$B_{\text{res}} = \sqrt{x} \, \text{mT}$$. Comparing, $$x = 20$$.

Hence, the required value is 20.