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Magnets have bewildered everyone from the Insane Clown Posse to Tim Allen. Stand apart from the madding crowd and set yourself straight on the fields of moving charges.

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A uniform magnetic field \(\vec{B}\) of magnitude \(1.5\text{ mT}\) is directed vertically upward. If a proton with kinetic energy \(5.5\text{ MeV}\) enters the chamber horizontally, what is the approximate magnetic deflecting force acting on the proton as it enters the magnetic field?

The proton mass is \(1.67 \times 10^{-27} \text{ kg}\) and \(eV=1.60 \times 10^{-19} \text{ J}.\) (Neglect Earth's magnetic field.)

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The above is a schematic diagram of a mass spectrometer. The magnitude of the electric field between the plates of the velocity selector is \(950 \text{ V/m},\) and the magnitude of the magnetic field (directed into the screen) in both the velocity selector and the deflection chamber is \(0.910 \text{ T}.\) If a singly charged ion with mass \(m=2.20 \times 10^{-25} \text{ kg}\) passes the velocity selector region with no deflection, what is the approximate radius of the path in deflection chamber?

The value of the elementary charge is \(e=1.60 \times 10^{-19} \text{ C}.\)

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In the above diagram, a charged particle with mass \(m=1.4 \times 10^{-15} \text{ kg}\) and charge \(q=2.6 \times 10^{-17} \text{ C}\) enters a region of uniform magnetic field \(B=0.8 \text{ T}\) (directed into the screen) and uniform electric field (directed downward). If the particle with a speed of \(v=15.0 \text{ m/s}\) passes through the device with no deflection, what is the magnitude of the electric field?

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A proton is traveling in a magnetic field of \(2.30 \text{ mT}\) with the angle of \(30.0^\circ\) with respect to the direction of the magnetic field. If the proton experiences a magnetic force of \(6.80 \times 10^{-17} \text{ N},\) what is the approximate kinetic energy of the proton?

The proton mass is \(1.67 \times 10^{-27} \text{ kg}\) and the value of the elementary charge is \(e=1.60 \times 10^{-19} \text{ C}.\)

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