Question

An electron moving horizontally with a velocity of 4 ✕ 10⁴ m/s enters a region of uniform magnetic field of 10⁻⁵ T acting vertically upward as shown in the figure. Draw its trajectory and find out the time it takes to come out of the region of magnetic 

field.

Answer 1

Let the time taken by the electron to come out of the region of magnetic field be t.
Velocity of the electron, v = 4 × 10⁴ m/s
Magnetic field, B = 10⁻⁵ T
Mass of the electron, m = 9 × 10⁻³¹ kg

We know

\[t = \frac{\pi r}{v}\]

\[\text { where r } = \frac{mv}{Bq}\]

\[\text { Now,} \]

\[t = \frac{\pi m}{Bq} = \frac{3 . 14 \times 9 \times {10}^{- 31}}{{10}^{- 5} \times 1 . 6 \times {10}^{- 19}}\]

\[ \Rightarrow t = 17 . 66 \times {10}^{- 7} s = 1 . 76 \mu s\]

Thus, the time taken by the electron to come out of the region of magnetic field is 1.76 μs.