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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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A metal sphere has a charge of +10C. What is the net charge after 8.0 x 10^13 electrons have been placed on it?
masya89 [10]

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After that, we have to calculate the charge given by 8.0*10^13 electrons, then we an additional charge of: 8.0*10^13 * -1,6*10^-19 C=1.28*10^-5C

Finally the net charge of the metal sphere, initially charged by +10C is:

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Xelga [282]

Answer:

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so the ratio for destructive interference is

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