An electron moves with a velocity of $1200 m\, s^{-1}$ in a magnetic field of 0.4 T at an angle of 30°. What will be the approximate radius of its path?

(Take e/m of electron = $1.76 × 10^{11} C\, kg^{-1}$)

$8.6 × 10^{-9} m$

The correct answer is Option (2) → $8.6 × 10^{-9} m$

Given:

Velocity, $v = 1200\ \text{m·s}^{-1}$

Magnetic field, $B = 0.4\ \text{T}$

Angle, $θ = 30°$

Charge on electron, $e = 1.6 \times 10^{-19}\ \text{C}$

Mass of electron, $m = 9.1 \times 10^{-31}\ \text{kg}$

Formula for radius of circular path:

$r = \frac{m v \sinθ}{e B}$

Substitute values:

$r = \frac{9.1 \times 10^{-31} \times 1200 \times \sin30°}{1.6 \times 10^{-19} \times 0.4}$

$r = \frac{9.1 \times 10^{-31} \times 1200 \times 0.5}{6.4 \times 10^{-20}}$

$r = \frac{5.46 \times 10^{-28}}{6.4 \times 10^{-20}} = 8.53 \times 10^{-9}\ \text{m}$

Answer: $r \approx 8.5 \times 10^{-9}\ \text{m}$