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)
Answer & explanation
Correct answer: option 3
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 $9.1 \times 10^{-4} \, T$.