F f f0 f 0 f 2 f0 abcde 2 why f qvb 2q 2v0

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Unformatted text preview: e of the particle is doubled (Q = 2q),while keeping the mass constant. How does the path of the particle change? – I expected no slam dunk.. Let’s talk about this one some more. BB How does v, the new velocity at the entrance, compare to the original velocity v0? v v0 v= 0 v = v0 (A) (B) (C) (D) (E) v= 2 2 v = 2v0 v = 2v0 • Why?? 12 12 mv = QEd = 2qEd = 2 mv0 2 2 2 v 2 = 2v0 v = 2v0 Follow­Up Follow­Up A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. q,m Assume q,m,E,d, and x0 are known. What is B? B= 2 x0 exits here x0/2 XXXXXXXXX X X X X X X X X X x0 XXXXXXXXX d XXXXXXXXX enters here B B 2mEd q E • Suppose the charge of the particle is doubled (Q = 2q),while keeping the mass constant. How does the path of the particle change? v = 2v0 How does F, the magnitude of the new force at the entrance, compare to F0, the magnitude of the original force? F F = F0 F= 0 F = 2 F0 (A) (B) (C) (D) (E) 2 • Why?? F = QvB = 2q ( ) 2v0 B F = 2 2 F0 F = 2 F0 F = 2 2 F0 BB Follow­Up Follow­Up A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region containing a constant magnetic field B at the points shown. q,m Assume q,m,E,d, and x0 are known. What is B? B= 2 x0 exits here x0/2 XXXXXXXXX X X X X X X X X X x0 XXXXXXXXX d XXXXXXXXX enters...
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