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A proton of charge e and mass m enters a...

A proton of charge `e` and mass `m` enters a uniform magnetic field `B = Bi` with an initial velocity `v=v_xhati+v_yhatj`. Find an expression in unit vector notation for its velocity at time `t`.

Text Solution

Verified by Experts

The correct Answer is:
`vecv(t)=v_(0x)hati+v_(0y)(sin omegat)hatk`
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Knowledge Check

  • An electron of charge -e , mass m, enters a uniform magnetic field vec(B)= B hat(i) with an initial velocity vec(v) = v_(x) hat(i) + v_(y) hat(j) . What is the velocity of the electron after a time interval of t seconds?

    A
    `v_(x) hat(i) + v_(y) hat(j) + (e )/(m) v_(y) Bt hat(k)`
    B
    `v_(x) hat(i) + v_(y) hat(j) - (e )/(m) v_(y) B t hat(k)`
    C
    `v_(x) hat(i) + (v_(y) + (e )/(m) v_(y) B t) hat(j)`
    D
    `(v_(x) + (e )/(m) v_(y) B t) hat(i) + v_(y) hat(j)`

  • A particle of charge -q and mass m enters a uniform magnetic field vecB (perpendicular to paper inward) at P with a velocity v_0 at an angle alpha and leaves the field at Q with velocity v at angle beta as shown in fig.

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    C
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    D
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