An electron with kinetic energy 5 eV enters a region of uniform magnetic field of 3 ?T perpendicular to its direction. An electric field E is applied perpendicular to the direction of velocity and magnetic field. The value of E, so that the electron moves

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Solution

Q.
An electron with kinetic energy 5 eV enters a region of uniform magnetic field of 3 μT perpendicular to its direction. An electric field E is applied perpendicular to the direction of velocity and magnetic field. The value of E, so that the electron moves along the same path, is ____ $N C^{- 1}$ . (Given, mass of electron $= 9 \times 10^{- 31}$ kg, electric charge $= 1.6 \times 10^{- 19}$ C) [2024]

Ans.
(4)   For the given condition of moving undeflected, net force should be zero.

   $q E = q V B \Rightarrow E = V B$

   $E = \sqrt{\frac{2 \times K E}{m}} \times B = \sqrt{\frac{2 \times 5 \times 1.6 \times 10^{- 19}}{9 \times 10^{- 31}}} \times 3 \times 10^{- 6} = 4 N / C$

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