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प्रश्न
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.
उत्तर
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.
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