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An electron of kinetic energy 1.59 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 35.4 cm. Find (a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.

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Answer:

a)
v = 2.36 \cdot 10^(7) m/s

b)
B = 3.80 \cdot 10^(-4) T

c)
f = 1.06 \cdot 10^(7) Hz

d)
T = 9.43 \cdot 10^(-8) s

Step-by-step explanation:

a) We can find the electron's speed by knowing the kinetic energy:


K = (1)/(2)mv^(2)

Where:

K: is the kinetic energy = 1.59 keV

m: is the electron's mass = 9.11x10⁻³¹ kg

v: is the speed =?


v = \sqrt{(2K)/(m)} = \sqrt{(2*1.59 \cdot 10^(3) eV*(1.602 \cdot 10^(-19) J)/(1 eV))/(9.11 \cdot 10^(-31) kg)} = 2.36 \cdot 10^(7) m/s

b) The electron's speed can be found by using Lorentz's equation:


F = q(v* B) = qvBsin(\theta) (1)

Where:

F: is the magnetic force

q: is the electron's charge = 1.6x10⁻¹⁹ C

θ: is the angle between the speed of the electron and the magnetic field = 90°

The magnetic force is also equal to:


F = ma_(c) = m(v^(2))/(r) (2)

By equating equation (2) with (1) and by solving for B, we have:


B = (mv)/(rq) = (9.11 \cdot 10^(-31) kg*2.36 \cdot 10^(7) m/s)/(0.354 m*1.6 \cdot 10^(-19) C) = 3.80 \cdot 10^(-4) T

c) The circling frequency is:


f = (1)/(T) = (\omega)/(2\pi) = (v)/(2\pi r)

Where:

T: is the period = 2π/ω

ω: is the angular speed = v/r


f = (v)/(2\pi r) = (2.36 \cdot 10^(7) m/s)/(2\pi*0.354 m) = 1.06 \cdot 10^(7) Hz

d) The period of the motion is:


T = (1)/(f) = (1)/(1.06 \cdot 10^(7) Hz) = 9.43 \cdot 10^(-8) s

I hope it helps you!

User Petok Lorand
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