An electron accelerated by a potential difference V=3.6 {V}, first enters a region of uniform electric field of a parallel-plate capacitor whose plates extend over a length l=6 {cm} in the direction of initial velocity. The electric field is normal to the direction of initial velocity and its strength varies with time as E=a \times t where a=3200 {Vm}^{-1} {s}^{-1}. Then the electron enters a region of uniform magnetic field of induction B=\pi \times 10^{-9} {T}. Direction of magnetic field is same as that of the electric field. Calculate the pitch (in {mm} ) of helical path traced by the electron in the magnetic field. (Mass of electron, m=9 \times 10^{-31} {kg}, e=1.6 \times 10^{-19} {C}) [Neglect the effect of induced magnetic field].

JEE Advanced 2020 Full Test 6 Paper 1

Question : 17 of 54

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An electron accelerated by a potential difference V=3.6V, first enters a region of uniform electric field of a parallel-plate capacitor whose plates extend over a length l=6cm in the direction of initial velocity. The electric field is normal to the direction of initial velocity and its strength varies with time as E=a×t where a=3200Vm−1s−1. Then the electron enters a region of uniform magnetic field of induction B=π×10−9T. Direction of magnetic field is same as that of the electric field. Calculate the pitch (in mm ) of helical path traced by the electron in the magnetic field. (Mass of electron, m=9×10−31kg,e=1.6×10−19C) [Neglect the effect of induced magnetic field].

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