Alpha particles (mass = 6.7 × 10^-27 kg, q = +2e) are accelerated from rest through a potential difference of 3.0 kV, then enter a uniform magnetic field B = 0.40 T perpendicular to their direction of motion. Calculate the radius of the path.

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Chapter1: Units, Trigonometry. And Vectors
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Alpha particles (mass = 6.7 × 10^-27 kg, q = +2e) are accelerated from rest through a potential difference of 3.0 kV, then enter a uniform magnetic field B = 0.40 T perpendicular to their direction of motion. Calculate the radius of the path.

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Alpha particles (mass = 6.7 × 10^-27 kg, q = +2e) are accelerated from rest through a potential difference of 3.0 kV. They then enter a uniform magnetic field B = 0.40 T perpendicular to their direction of motion. Calculate the radius of their path.

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We know that, sadius of charge
moving in magnetic field,
R= mu
98
why
not use this eft.?
It can be modified as.
2 mq v
R=
√2mV
9 B
Put the known values,
R = √2 (6.7x10-27) (3x10³)
√2x4.6x10-19) (0.40)
R = 4.483 X 10-12
=
4x10-10 x0.40
R=
јава
I don't understand
Why use this?
0.0280 m
Answer
Transcribed Image Text:solution We know that, sadius of charge moving in magnetic field, R= mu 98 why not use this eft.? It can be modified as. 2 mq v R= √2mV 9 B Put the known values, R = √2 (6.7x10-27) (3x10³) √2x4.6x10-19) (0.40) R = 4.483 X 10-12 = 4x10-10 x0.40 R= јава I don't understand Why use this? 0.0280 m Answer
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