1. Magnetic field at the center of a circular coil

For a coil of $N$ turns, radius $R$, carrying current $I$,

$B=\frac{\mu_0 N I}{2R}$

2. Given data

• Number of turns in each coil: $N=100$
• Radius: $R=\pi\,\text{cm}=\pi\times 10^{-2}\,\text{m}$
• Current in coil $P$: $I_1=1\,\text{A}$
• Current in coil $Q$: $I_2=2\,\text{A}$
• $\mu_0=4\pi\times 10^{-7}\,\text{T m A}^{-1}$

3. Field due to coil $P$

$B_1=\frac{\mu_0 N I_1}{2R}$

Substitute:

$B_1=\frac{(4\pi\times 10^{-7})(100)(1)}{2(\pi\times 10^{-2})}$

Cancel $\pi$:

$B_1=\frac{4\times 10^{-5}}{2\times 10^{-2}}=2\times 10^{-3}\,\text{T}=2\,\text{mT}$

4. Field due to coil $Q$

$B_2=\frac{\mu_0 N I_2}{2R}$

Since $I_2=2\,\text{A}$,

$B_2=2B_1=4\,\text{mT}$

5. Resultant field

The planes of the coils are mutually perpendicular, so their axes are also mutually perpendicular. Hence the magnetic fields at the common center are perpendicular.

Therefore,

$B=\sqrt{B_1^2+B_2^2}$

$B=\sqrt{(2)^2+(4)^2}\,\text{mT}$

$B=\sqrt{4+16}\,\text{mT}=\sqrt{20}\,\text{mT}$

Comparing with $\sqrt{x}\,\text{mT}$,

$x=20$

6. Comparison with stored answer

Stored correct answer = $20$.

This matches the derived answer.