A circular coil with 50 turns and a radius of 2.0 cm carries a 5.0...

The maximum torque on a current-carrying coil in a magnetic field is given by the formula:

\tau_{\text\max} = NIAB

where:
$N$ is the number of turns $= 50$
$I$ is the current $= 5.0 \text{ A}$
$A$ is the area of the coil $= \pi r^2 = \pi (2.0 \times 10^{-2} \text{ m})^2 = 4\pi \times 10^{-4} \text{ m}^2$
$B$ is the magnetic field $= 0.20 \text{ T}$

Substituting the given values:

\tau_{\text\max} = 50 \times 5.0 \times (4\pi \times 10^{-4}) \times 0.20

\tau_{\text\max} = 250 \times 0.20 \times 4\pi \times 10^{-4}

\tau_{\text\max} = 50 \times 4\pi \times 10^{-4}

\tau_{\text\max} = 200\pi \times 10^{-4}

\tau_{\text\max} = 2\pi \times 10^{-2} \approx 2 \times 3.14 \times 10^{-2} = 6.28 \times 10^{-2} \text{ N m}

Rounding to two significant figures, we get $6.3 \times 10^{-2} \text{ N m}$.