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Unformatted text preview: ou may assume that the electrons move at right angles to the magnetic field. Physics 7B Chargetomass: e/m p. 2 2. Recall from electrostatics, earlier in the course, that an electron obtains kinetic energy when accelerated across a potential difference V. Since we can directly measure the accelerating voltage V in this experiment, but not the electrons’ velocity v, replace velocity in your previous equation with an expression containing voltage. The electron starts at rest. (Don’t get capital V, voltage, confused with lowercase v, velocity.) Now solve this equation for e/m. You should obtain e 2V = 22 m Br Eq. 1 Physics 7B Chargetomass: e/m p. 3 3. The magnetic field on the axis of a circular current loop a distance z away is given by B= µ 0 IR2 2( R + z
2 3 22 , Eq. 2 ) where R is the radius of the loop and I is the current. (See example in text for a derivation and discussion of this result.) Using this result, calculate the magnetic field at the midpoint along the axis between the centers of the two current loops that make up the Helmholtz coils, in terms
(Fig. from D. Giancoli’s Physics) of their number of turns N, current I, and radius R—see Fig. 2 on p...
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 Fall '10
 yildiz
 Physics, Charge, Magnetism, Mass

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