E2HWCH26 - Chapter 26 The Magnetic Field 1[SSM When the...

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Chapter 26 The Magnetic Field 1 [SSM] When the axis of a cathode-ray 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 26-30. 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 right-hand 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 right-hand 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 . 0 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 . 0 ˆ T 75 . 0 ˆ m/s 10 75 . 2 nC 64 . 3 3 μ - = + × × - = ( c ) Evaluate F for B = 0.65 T i ˆ : ( 29 ( 29 ( 29 [ ] 0 ˆ T 65 . 0 ˆ m/s 10 75 . 2 nC 64 . 3 3 = × × - = i i F
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( 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 . 0 ˆ T 75 . 0 ˆ m/s 10 75 . 2 nC 64 . 3 3 μ = + × × - = 18 A straight segment of a current-carrying wire has a current element L I , where I = 2.7 A and j i L ˆ cm 0 . 4 ˆ cm 0 . 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 . 0 ˆ T 3 . 1 ˆ cm 0 . 4 ˆ cm 3.0 A 7 . 2 - = × + = 27 [SSM] A proton moves in a 65-cm-radius circular orbit that is perpendicular to a uniform magnetic field of magnitude 0.75 T. ( a ) What is the orbital period for the motion? ( b ) What is the speed of the proton? ( c ) What is the kinetic energy of the proton?
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