Asked by Robert
An antiproton (which has the same properties as a proton except that its charge is -e) is moving in the combined electric and magnetic fields of the figure: h tt p://post image. org/image/2sh3s3hs4/
What are the magnitude and direction of the antiproton's acceleration at this instant?
What would be the magnitude and direction of the acceleration if \vec v were reversed?
What are the magnitude and direction of the antiproton's acceleration at this instant?
What would be the magnitude and direction of the acceleration if \vec v were reversed?
Answers
Answered by
drwls
Correct on the negative charge of the antiproton. The magnetic force is q V x B (vector cross product) and the electric force is q E (scalar multiplication of E by a constant).
The magnetic force is in the direction of the red arrows (from the right-hand rule, with negative q), and the electric force is opposite to that.
If the direction of V is reversed, the electric force remaqins the same and the directikon of the magnetic force reverses.
The magnetic force is in the direction of the red arrows (from the right-hand rule, with negative q), and the electric force is opposite to that.
If the direction of V is reversed, the electric force remaqins the same and the directikon of the magnetic force reverses.
Answered by
Robert
Solved it. Thanks.
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