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encounters a medium with uniform magnetic flux density B - (33 - 22) T. Find the initial acceleration of the electron at the moment it enters the medium (given e1.6 x 10-19 C and me 9.1 x 10-31 kg).

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Answer #1

dynamics of an electron entering an uniform magnetic field is given as  u = Ve * x(positivedir ection)

Ve = 2 * 10^{6} meter/second

Me = 9.1 * 10^{-31} kg e = 1.6 * 10^{-19} c

to find intial acceleration vector of an electron

1. megnetic force

Fm = q U x B

2. newton second law

F = m * a

a = F / m = q U x B / m = -1.6 * 10^{-19} c * - x 2 * 10^{6} x (X3 - Z2) /  9.1 *10-31c

a = y * -1.6 * 10^{-19} * 2 *10^{6} / 9.1 * 10^{-31}

a = y .-1.05 * 10^{18} meter/second^{2} (acceleration in negetive Z direction)

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