An electron travels along a circle of radius 1 m in a perpendicular...

The radius of the circular path of a charged particle in a perpendicular magnetic field is given by $r = \dfrac{mv}{qB}$.

Rearranging for speed $v$, we get $v = \dfrac{qBr}{m}$.

Substituting the given values:

$v = \dfrac{1.6 \times 10^{-19} \times 0.50 \times 1}{9.1 \times 10^{-31}}$

$v = \dfrac{0.8 \times 10^{-19}}{9.1 \times 10^{-31}} \approx 8.8 \times 10^{10} \text{ m s}^{-1}$

The calculated speed is $8.8 \times 10^{10} \text{ m s}^{-1}$, which is significantly greater than the speed of light in vacuum ($c = 3 \times 10^{8} \text{ m s}^{-1}$).

Since no material particle can travel faster than the speed of light, this speed is unreasonable.