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Unformatted text preview: Solution Derivations for Capa #8 1) A mass spectrometer applies a voltage of 2 . 00 kV to accelerate a singly charged ion (+ e ). A 0 . 400 T field then bends the ion into a circular path of radius 0 . 305 m . What is the mass of the ion? V = Given q = Given B = Given r = Given A mass spectrometer will cause the particle to follow a circular pattern. Thus, we know that F = ma = m v 2 r = qvB where the last step follows as a result of the magnetic field. To find the mass of the ion, we can solve for m . m v 2 r = qvB m = qvBr v 2 = qBr v Now the only problem is to find r . It is related to the kinetic energy gained by accelerating though the electric field. Since it is initially at rest, and exits the electric field before entering the mass spectrometer, the final kinetic energy will be the potential energy gained by traveling though the electric field. That is, 1 2 mv 2 = qV v = 2 qV m Plugging v into the above equation and squaring both sides, m = qBr 2 qV m m 2 = q 2 B 2 r 2 2 qV m = mqB 2 r 2 2 V m = qB 2 r 2 2 V 2) A charged particle is moving perpendicularly to a magnetic field B . Fill in the blank indicating the direction for the quantity missing in the table. (If your three answers from top to bottom are +z, x, y, then enter: +zxy in the computer). Use the diagram on the upper left for the directions of the various axes. 1 Use the right hand rule. Remember that if the charge is negative, it will move in the direction opposite that given by the right hand rule. 3) The rectangular loop shown below is located in a uniform magnetic field of 0 . 50 T pointing in the positive x direction. What is the magnitude of the magnetic dipole moment of the loop? DATA: h = 6 . 5 cm , w = 8 . cm , = 36 ....
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This note was uploaded on 03/24/2011 for the course PHYC 1100Y taught by Professor Thomasj.duck during the Spring '11 term at Dalhousie.
 Spring '11
 THOMASJ.DUCK

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