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N76 [4]
4 years ago
5

A charged particle enters into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vec

tor. Ignoring the particle's weight, what type of path will the particle follow?
The charged particle will follow a straight-line path.
The charged particle will follow a spiral path.
The charged particle will follow a parabolic path.
The charged particle will follow a circular path.
Physics
1 answer:
Anit [1.1K]4 years ago
8 0

Answer:

The charged particle will follow a circular path.

Explanation:

The magnetic force exerted on the charged particle due to the magnetic field is given by:

F=qvB sin \theta

where

q is the charge

v is the velocity of the particle

B is the magnetic field

\theta is the angle between v and B

In this problem, the velocity is perpendicular to the magnetic field, so \theta = 90^{\circ}, sin \theta=1 and the force is simply

F=qvB

Moreover, the force is perpendicular to both B and v, according to the right-hand rule. Therefore, we have:

- a force that is always perpendicular to the velocity, v

- a force which is constant in magnitude (because the magnitude of v or B does not change)

--> this means that the force acts as a centripetal force, so it will keep the charged particle in a uniform circular motion. So, the correct answer is

The charged particle will follow a circular path.

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Answer:

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Explanation:

given,

mass of water = 2000 grams

initial temperature = 0° C

Final temperature = 100° C

specific heat of water (c) = 4.186 joule/gram

energy = m c Δ T                            

            = 2000 × 4.186 × (100° - 0°)

            = 837200 J                          

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hence, the stove energy went into heating water is 837.2 kJ.

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Answer:

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Explanation :

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According to Snell's law :

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