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mixas84 [53]
4 years ago
5

A charged particle is observed traveling in a circular path of radius R in a uniform magnetic field. If the particle were travel

ing twice as fast, the radius of the circular path would be:
Physics
1 answer:
Jobisdone [24]4 years ago
3 0

Answer:

radius of circular path becomes doubled

Explanation:

As the charged particle is moving in a magnetic field then it experiences a centripetal force.

So, the magnetic force is balanced by the centripetal force

q\times v\times B = \frac{mv^{2}}{R}

where, m be the mass of charged particle, q be the charge and v be the velocity, B be the magnetic field, R be the radius of circular path.

R = \frac{mv}}{qB}

Here we observe that the radius of the path is directly proportional to the velocity of the charged particle.

R ∝ v    ..... (1)

let the new radius be R' as the velocity is doubled

R' ∝ 2 v    ..... (2)

Divide equation (2) by equation (1) we get

R' = 2 R

Thus, the radius of circular path becomes doubled.

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A cue ball initially moving at 3.4 m/s strikes a stationary eight ball of the same size and mass. After the collision, the cue b
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Answer:

speed of eight ball speed after the collision is 3.27 m/s

Explanation:

given data

initially moving v1i = 3.4 m/s

final speed is v1f = 0.94 m/s

angle = θ w.r.t. original line of motion

solution

we assume elastic collision

so here using conservation of energy

initial kinetic energy = final kinetic energy .............1

before collision kinetic energy = 0.5 × m× (v1i)²

and

after collision kinetic energy =  0.5 × m× (v1f)²  + 0.5 × m× (v2f)²

put in equation 1

0.5 × m× (v1i)² =  0.5 × m× (v1f)²  + 0.5 × m× (v2f)²

(v2f)² = (v1i)² - (v1f)²

(v2f)² = 3.4² - 0.94²

(v2f)² = 10.68

taking the square root both

v2f = 3.27 m/s

speed of eight ball speed after the collision is 3.27 m/s

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