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Unformatted text preview: Chapter 26 The Magnetic Field 1 • [SSM] When the axis of a cathoderay tube is horizontal in a region in which there is a magnetic field that is directed vertically upward, the electrons emitted from the cathode follow one of the dashed paths to the face of the tube in Figure 2630. The correct path is ( a ) 1, ( b ) 2, ( c ) 3, ( d ) 4, ( e ) 5. Determine the Concept Because the electrons are initially moving at 90 ° to the magnetic field, they will be deflected in the direction of the magnetic force acting on them. Use the righthand rule based on the expression for the magnetic force acting on a moving charge B v F × = q , remembering that, for a negative charge, the force is in the direction opposite that indicated by the righthand rule, to convince yourself that the particle will follow the path whose terminal point on the screen is 2. ) ( b is correct. 15 • A point particle has a charge equal to –3.64 nC and a velocity equal to i m/s 10 75 . 2 3 × . Find the force on the charge if the magnetic field is ( a ) 0.38 T ˆ j , ( b ) 0.75 T ˆ i + 0.75 T ˆ j , ( c ) 0.65 T ˆ i , and ( d ) 0.75 T ˆ i + 0.75 T ˆ k . Picture the Problem The magnetic force acting on the charge is given by B v F × = q . We can express v and B , form their vector (also known as the ″ cross ″ ) product, and multiply by the scalar q to find F . Express the force acting on the charge: B v F × = q Substitute numerical values to obtain: ( 29 ( 29 [ ] B i F × × = ˆ m/s 10 75 . 2 nC 64 . 3 3 ( a ) Evaluate F for B = 0.38 T j ˆ : ( 29 ( 29 [ ( 29 ] ( 29 k j i F ˆ N 8 . 3 ˆ T 38 . ˆ m/s 10 75 . 2 nC 64 . 3 3 μ = × × = ( b ) Evaluate F for B = 0.75 T i ˆ + 0.75 T j ˆ : ( 29 ( 29 [ ( 29 ( 29 { }] ( 29 k j i i F ˆ N 5 . 7 ˆ T 75 . ˆ T 75 . ˆ m/s 10 75 . 2 nC 64 . 3 3 μ = + × × = ( c ) Evaluate F for B = 0.65 T i ˆ : ( 29 ( 29 ( 29 [ ] ˆ T 65 . ˆ m/s 10 75 . 2 nC 64 . 3 3 = × × = i i F ( d ) Evaluate F for B = 0.75 T i ˆ + 0.75 T k ˆ : ( 29 ( 29 [ ( 29 ( 29 ] ( 29 j k i i F ˆ N 5 . 7 ˆ T 75 . ˆ T 75 . ˆ m/s 10 75 . 2 nC 64 . 3 3 μ = + × × = 18 • A straight segment of a currentcarrying wire has a current element L I , where I = 2.7 A and j i L ˆ cm . 4 ˆ cm . 3 + = . The segment is in a region with a uniform magnetic field given by i ˆ T 3 . 1 . Find the force on the segment of wire. Picture the Problem We can use B L F × = I to find the force acting on the wire segment. Express the force acting on the wire segment: B L F × = I Substitute numerical values and evaluate F : ( 29 ( 29 ( 29 [ ] ( 29 ( 29 k i j i F ˆ N 14 ....
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This note was uploaded on 03/07/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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