Two circular coils having radii of 5.0 cm and 10 cm carry equal...

The magnetic field at the centre of a circular coil of $N$ turns, radius $R$, carrying current $I$ is given by $B = \dfrac{\mu_0 N I}{2 R}$.

For the inner coil:
$B_1 = \dfrac{4\pi \times 10^{-7} \times 50 \times 2.0}{2 \times 5.0 \times 10^{-2}} = 4\pi \times 10^{-4} \text{ T}$

For the outer coil:
$B_2 = \dfrac{4\pi \times 10^{-7} \times 100 \times 2.0}{2 \times 10 \times 10^{-2}} = 4\pi \times 10^{-4} \text{ T}$

When the outer coil is rotated by 90^{}^{{}^{\circ}} about its diameter, its plane becomes perpendicular to the plane of the inner coil. Consequently, their magnetic fields at the centre will be mutually perpendicular.

The resultant magnetic field $B$ is:
$B = \sqrt{B_1^2 + B_2^2}$

$B = \sqrt{(4\pi \times 10^{-4})^2 + (4\pi \times 10^{-4})^2}$

$B = 4\pi \sqrt{2} \times 10^{-4} \text{ T}$

$B \approx 4 \times 3.14 \times 1.414 \times 10^{-4} \text{ T} \approx 1.77 \times 10^{-3} \text{ T}$

Rounding to one decimal place, $B \approx 1.8 \times 10^{-3} \text{ T}$.