An electron of mass $9.1×10^{-31} kg$ carrying charge $1.6 × 10^{-19} C$, is moving horizontally with a speed of $4×10^6 m/s$ in a uniform magnetic field of magnitude $8×10^{-4} T$ acting vertically upward. The radius of its circular path is,

2.84 cm

The correct answer is Option (2) → 2.84 cm

Radius of circular path in magnetic field:

$r = \frac{mv}{qB}$

Given: $m = 9.1 \times 10^{-31} \ \text{kg}$, $v = 4 \times 10^6 \ \text{m/s}$, $q = 1.6 \times 10^{-19} \ \text{C}$, $B = 8 \times 10^{-4} \ \text{T}$

$r = \frac{(9.1 \times 10^{-31})(4 \times 10^6)}{(1.6 \times 10^{-19})(8 \times 10^{-4})}$

$r = \frac{36.4 \times 10^{-25}}{1.28 \times 10^{-23}}$

$r \approx 2.84375 \times 10^{-2} \ \text{m}$

Radius of the circular path ≈ 2.84 cm