An electron having 1.14 kev of kinetic energy is moving in a · region of uniform

Question

An electron having 1.14 kev of kinetic energy is moving in a · region of uniform magnetic field of strength 21.3 mT as shown in the drawing. (Note the mass of an electron is 9.11x10 kg) ? ? ? ? a) Determine the direction that the electron is deflected. (left or right in the drawing) b) Determine the radius of curvature of the electron's circular path in the magnetic field Ifthe electron stays in the magnetic field, determine the period of its orbit within the magnetic field. (cyclotron period) c)

Explanation / Answer

(A) F = q (v x B)

q = - ve

v = j^

B = -k^

F = (-1)(j^ x -k^) = i^

Ans: to the right

(B) Fb = m a_c

q v B = m v^2 / r

r = m v / q B

{ KE = m v^2 /2

1.14 x 10^6 x 1.6 x 10^-19 = (9.11 x 10^-31)v^2 /2

v = 2 x 10^7 m/s}

r = (9.11 x 10^-31)(2 x 10^7)/(1.6 x 10^-19)(0.0213)

r = 5.35 x 10^-3 m Or 5.35 mm

(c) T = 2 pi r / v

T = 1.68 x 10^-9 sec

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