An electron is moving at a velocity of $3.2 × 10^7 m/s$ in a magnetic field along a circular path of radius 20 cm. The magnitude of the magnetic field applied is (approximately)

$9 × 10^{-4} T$

The correct answer is Option (3) → $9 × 10^{-4} T$

Given:

Velocity of electron, $v = 3.2 \times 10^7 \, m/s$

Radius of circular path, $r = 20 \, cm = 0.2 \, m$

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

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

Formula:

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

or, $B = \frac{mv}{er}$

Substitute:

$B = \frac{9.1 \times 10^{-31} \times 3.2 \times 10^7}{1.6 \times 10^{-19} \times 0.2}$

$B = \frac{2.912 \times 10^{-23}}{3.2 \times 10^{-20}}$

$B = 9.1 \times 10^{-4} \, T$

∴ The magnitude of the magnetic field is approximately $9.1 \times 10^{-4} \, T$.