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Electric potential difference to accelerate electron

  • Thread starter jegues
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  • #1
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Homework Statement



What electric potential difference is required to accelerate and electron, initially at rest, to the speed at which it will move in a circular path with radius 1cm when the electron enters a region of uniform magnetic field of 1.0T, if the field is perpendicular to the electron's velocity? (Ignore gravity)

Homework Equations





The Attempt at a Solution



We know that,

[tex]F_{B} = qvB = \frac{mv^{2}}{r}[/tex]

So,

[tex]v = \frac{qBr}{m}[/tex]

The part I'm confused about is how to relate this knowledge the the potential difference?

Any suggestions?

EDIT: I figured it out.

[tex]\Delta V = \frac{qB^{2}r^{2}}{2m}[/tex]
 

Answers and Replies

  • #2
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change in the potential energy is equal to gain in kinetic energy

qV = 0.5mv2
 

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