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**Unformatted text preview: **homework 18 - LEE, BENJAMIN - Due: Oct 16 2007, 4:00 am 1 Question 1, chap 30, sect 1. part 1 of 3 10 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 direction) and a 0 . 2 T uniform mag- netic field (aligned with the k-direction) as shown in the figure by the shaded area. q m Region of Magnetic Field . 2 T +2800 V 1 cm x y z 3 9 c m Region I Region II Figure: is in the direction + x (to the right), is in the direction + y (up the page), and k is in the direction + z (out of the page). In which direction (relative to the co- ordinate system shown above) should the magnetic field point in order for negatively charged ions to move along the path shown by the dotted line in the diagram above? 1. vector B bardbl vector B bardbl = k correct 2. bardbl vector B bardbl = 0 ; direction undetermined 3. vector B bardbl vector B bardbl = + 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 positive ions to move in the semicircle. As the negatively 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 or in the z-direction k parenrightBig , since the force vector F is in the direction down the page; i.e. , " " q | q | = vectorv bardbl vectorv bardbl = + vector B bardbl vector B bardbl = ? vector F B bardbl vector F B bardbl = , and the vector product k = , and since vector F = qvectorv vector B = bardbl vector F bardbl ( ) vector F bardbl vector F bardbl = q | q | bracketleftBigg vectorv bardbl vectorv bardbl vector B bardbl vector B bardbl bracketrightBigg = bracketleftBig (+ ) parenleftBig k parenrightBigbracketrightBig = , consequently vector B bardbl vector B bardbl = k is correct . Question 2, chap 30, sect 1. part 2 of 3 10 points In "Region I", the electric potential be- tween the plates is 2800 V, the distance be- tween the plates is 1 cm, and the magnetic field in both "Regions I and II" is 0 . 2 T ....

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