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oldhw 20 - nguyen(jmn727 oldhomework 20 Turner(59070 This...

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nguyen (jmn727) – oldhomework 20 – Turner – (59070) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A device (“source”) emits a bunch of charged ions (particles) with a range of ve- locities (see figure). Some of these ions pass through the left slit and enter “Region I” in which there is a vertical uniform electric field (in the ˆ direction) and a 0 . 3 T uniform mag- netic field (aligned with the ± ˆ k -direction) as shown in the figure by the shaded area. q m Region of Magnetic Field 0 . 3 T +2400 V 0 . 6 cm x y z 36 cm Region I Region II Figure: ˆ ı is in the direction + x (to the right), ˆ is in the direction + y (up the page), and ˆ k is in the direction + z (out of the page). In which direction (relative to the coor- dinate system shown above) should the mag- netic field point in order for positively charged ions to move along the path shown by the dot- ted line in the diagram above? 1. vector B bardbl vector B bardbl = + ˆ k correct 2. bardbl vector B bardbl = 0 ; direction undetermined 3. vector B bardbl vector B bardbl = ˆ k Explanation: To obtain a straight orbit, the upward and downward forces need to cancel. The force on a charged particle is vector F = vector F E + vector F B = q ( vector E + vectorv × vector B ) . For the force to be zero, we need vector F E + vector F B = 0 , or vector F E = vector F B . Therefore, the forces are equal and opposite and the magnitude of forces are equal; i.e. , bardbl vector F E bardbl = bardbl vector F B bardbl . The force due to the magnetic field provides the centripetal force that causes the negative ions to move in the semicircle. As the positively charged ion exits the re- gion of the electric field, vector F B = q vectorv × vector B , so by the right-hand rule the magnetic field must point out of the page parenleftBig
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