Results 1 to 5 of 5

Math Help - Magnetic fields and motor principle

  1. #1
    Member
    Joined
    Dec 2007
    Posts
    76

    Magnetic fields and motor principle

    1) An electron moves at a speed of 5.0x10^6m/s perpedicular to a uniform magnetic field. The path of the electron is a circle of radius 1.0x10^-3m.

    a)what is the magnitude of the magnetic field?

    b)what is the magnitude of the electron's acceleraton in the field


    2) A beam of protons move in a circle of radius 0.22m at right angles to a 0.35T magnetic field.

    a)What is the speed of each proton?

    b)what is the magnitude of the centripetal force acting on each proton.

    Tried a few equations but still can't seem to get the right answers....
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,940
    Thanks
    338
    Awards
    1
    Quote Originally Posted by johett View Post
    1) An electron moves at a speed of 5.0x10^6m/s perpedicular to a uniform magnetic field. The path of the electron is a circle of radius 1.0x10^-3m.

    a)what is the magnitude of the magnetic field?

    b)what is the magnitude of the electron's acceleraton in the field
    The electron is moving at a constant speed in a circular path. So applying Newton's second law in the radial direction (+r toward the center) at a particular instant, we get
    \sum F_r = F_{mag} = F_C = \frac{mv^2}{r}
    where
    F_{mag} = qvB

    So
    qvB = \frac{mv^2}{r}

    B = \frac{mv}{qr}
    (This is the "cyclotron formula" rewritten a bit.)

    To get the acceleration of the electron, notice that the acceleration will simply be the centripetal acceleration, so
    a = \frac{v^2}{r}

    Problem 2 is almost identical to this one.

    -Dan
    Last edited by topsquark; March 30th 2008 at 10:18 AM. Reason: Made an oopsie!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2007
    Posts
    76
    B= \frac{mv}{qvr}

    = (9.11x10^-31)(5.0x10^6m/s) / (1.602x10^-19)(5.0x10^6m/s)(1.0x10^-3)
    =5.0x10^-9

    I still can't seem to get the right answer... I used 1.602x10^-19 for q and 9.11x10^-31 for the mass of the electron. The answer is suppose to be 2.8x10^-2 T.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2008
    Posts
    10
    Topsquark made a little mistake, which you should have noticed when entering the formula in your calculator.

    The correct formula for B should be:
    B= \frac{mv}{qr}

    This will yield B = 0,0284  T or B = 2,84 \times 10^{-2} T
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,940
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Flippy View Post
    Topsquark made a little mistake, which you should have noticed when entering the formula in your calculator.

    The correct formula for B should be:
    B= \frac{mv}{qr}

    This will yield B = 0,0284  T or B = 2,84 \times 10^{-2} T
    Thanks for the catch. I have fixed it in my original post.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Magnetic fields and electrical current.
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: June 24th 2010, 07:34 AM
  2. Replies: 1
    Last Post: April 9th 2010, 07:55 AM
  3. Electrical motor question
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: January 22nd 2009, 06:28 AM
  4. motor principle
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: March 30th 2008, 01:01 PM
  5. Replies: 2
    Last Post: December 2nd 2006, 08:07 AM

Search Tags


/mathhelpforum @mathhelpforum