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Archy [21]
3 years ago
8

1. A beam of charged particles enters a magnetic field and travels in a helical or spiral path What does the shape of the path i

ndicate about the initial velocity of the particles?
A. The initial velocity is parallel to the magnetic field.
B. The initial velocity has components both parallel and perpendicular to the magnetic field.
C. The initial velocity is perpendicular to the magnetic field.
D. The initial velocity has greater magnitude than particles traveling in a circle through the magnetic field.
Physics
1 answer:
Makovka662 [10]3 years ago
6 0

B. The initial velocity has components both parallel and perpendicular to the magnetic field.

Explanation:

When a charged particle moves in a magnetic field, it experiences a force which is given by

F=qvB sin \theta

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

\theta is the angle between the directions of v and B

From this formula, it is clear that the force has only a component in the direction perpendicular to direction of the the motion of the particle. Therefore:

  • if the initial velocity is perpendicular to the magnetic field, then the particle will follow a perfectly circular path
  • If the initial velocity is parallel to the magnetic field, then the particle will continue moving straight, since it will experience  no force
  • If the initial velocity has components both parallel and perpendicular to the magnetic field, then it will move in a helical or spiral path, which is a composition of the two motions above (straight motion + circular motion)
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n200080 [17]

Answer:

a. t = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}  b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0}  +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

 0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁  = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

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