An electron (mass $9 × 10^{-31} kg$ and charge $1.6 × 10^{-19} C$) moving with a speed of $5.0×10^7\, m/s$ enters a region of uniform magnetic field of strength 0.04 T which is perpendicular to its velocity. The radius of the circular path it follows is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $7.03 × 10^{-3} m$
The radius of the circular path of a charged particle moving perpendicular to a magnetic field is:
$r = \frac{mv}{qB}$
Given: $m = 9 \times 10^{-31} \, \text{kg}$, $v = 5.0 \times 10^7 \, \text{m/s}$, $q = 1.6 \times 10^{-19} \, \text{C}$, $B = 0.04 \, \text{T}$
Substitute values:
$r = \frac{9 \times 10^{-31} \cdot 5.0 \times 10^7}{1.6 \times 10^{-19} \cdot 0.04}$
$r = \frac{4.5 \times 10^{-23}}{6.4 \times 10^{-21}}$
$r \approx 7.03 \times 10^{-3} \, \text{m}$
Radius of circular path ≈ 7.0 mm