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On electron accelerated by a potential difference V=3 volt first enters into a uniform electric field of a parallel plate capacitor whose plates extend over a length $\mathrm{I}=6 \mathrm{~cm}$ in the direction of initial velocity. The electric field is normal to the direction of initial velocity and its strength varies with times as $\mathrm{\mathrm{E}=\alpha \mathrm{t}}$ , where $\mathrm{\alpha=3600 \mathrm{Vm}-1 \mathrm{~S}-1}$ . Then the electron enters into a uniform magnetic field of induction $\mathrm{B=\pi \times 10-9 \mathrm{~T}}$ . Direction of magnetic field is same as that of the electric field. The pitch of helical path traced by the electron in the magnetic field. (Mass of electron, $\mathrm{m=9 \times 10-31 \mathrm{~kg}}$ )
Option: 1
10.225 cm

Option: 2
5.215 cm

Option: 3
2.225 cm

Option: 4
1.215 cm

Answers (1)

best_answer

Pitch = time period $\times$ V(parallel)

Posted by

shivangi.shekhar

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