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galben [10]
2 years ago
9

An electron is accelerated through 2.40×10³V from rest and then enters a uniform 1.70-T magnetic field. What are(b) the minimum

values of the magnetic force this particle experiences?
Physics
1 answer:
Korolek [52]2 years ago
5 0

The maximum value of Force is 7.88x10^{-12}

The minimum value of magnetic force is 0

Potential V=2.40x10^{3}.

Magnetic field = B = 1.70 T.

we need to find

a) The maximum value of Force

Since the particle is moving from rest, the potential energy change is 2.40x10^{3}?

Potential Energy= vq= 2,40x10^{3}x (-1.6 x 10x10^{-19}%).

= -8.84x10^{-16}

Now this will be equal to 1/2 mv²

=  1/2 mv²= 8.84x 10^{-16}.

v=2.903x10^{7}m/s :

Wow Fmax = qvb

=1.6x10^{-19}x2.903x10^{7}x1.70 .

=7.88x10^{-12}

b) The minimum value of magnetic force

The minimum Value of magnetic force will be

when sinθ=0

The minimum value is 0

To know more about magnetic field refer to brainly.com/question/13091447

#SPJ4

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Explanation:

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If the electron wants to jump from the first energy level, n = 1, to the second energy level n = 2. The second energy level has higher energy than the first, so to move from n = 1 to n = 2, the electron needs to gain energy. It needs to gain (-3.4) - (-13.6) = 10.2 eV of energy to be excited to the second energy level.

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None of these transitions in the hydrogen atom corresponds to a photon energy of 5eV hence no photon of this energy is absorbed or emitted by the hydrogen atom.

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