An electron beam enters a crossed-field velocity selector with magnetic and electric fields of 2.0 mT and 6.0×10^3 N/C, respecti
1 answer:
Answer:
a) v = 3 10⁶ m / s , b) a = 1.055 10¹² m / s² , c) r = 8.53 m
Explanation:
a) Let's use Newton's Second Law of Balance, so that electrons do not deviate
-
= 0
=
q E = q v B
v = E / B
Let's calculate
v = 6.0 10³ / 2.0 10⁻³
v = 3 10⁶ m / s
b) If the electric field is disconnected, the only force left is the magnetic one
= m a
q v B = m a
a = q / m v B
a = q / m (E / B) B
a = q / m E
a = 1.6 10⁻¹⁹ /9.1 10⁻³¹ 6.0 10³
a = 1.055 10¹² m / s²
c) Acceleration is centripetal
a = v² / r
r = v² / a
r = (3 10⁶)² /1.055 10¹²
r = 8.53 m
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(1)
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Answer:
μ = 0.692
Explanation:
In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.
Attached is an image with the respective forces:
A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.
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