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**Unformatted text preview: **Physics 21 Fall, 2011 Solution to HW-10 27-3 In a 1 . 35 T magnetic field directed vertically upward, a particle having a charge of magnitude 8 . 90 C and initially moving northward at 4 . 72 km / s is deflected toward the east. (a) What is the sign of the charge of this particle? (b) Find the magnetic force on the particle. The magnetic force on a charged particle moving in a mag- netic field is given by the equation F = q v B Since, in this case, F , v , and B are mutually perpendicular, the magnitude of F is simply given by F = qvB , with the direction determined by the right hand rule. (a) We need to apply the right hand rule to see if the di- rection of the force is consistent with a positive charge or a negative charge. Imagine you are seated so that north is in front of you. The other directions are then determined: east is to the right, south is behind you, and west is to the left. So northward velocity means the particle is moving forward. Point the fingers of your right hand straight forward. The magnetic field is upward, so curl the fingers of your right hand upward. In order to do this, your palm must be facing upward. Then, the thumb of your right hand is pointing to the right (eastward). Eastward is the direction the particle is deflected. Thus, the particle must have a positive charge. Here is a diagram: N E S W B v F (b) F = qvB = (8 . 90 C)(4 . 72 km / s)(1 . 35 T) = 0 . 0567 N 27-4 A particle with mass m = 1 . 81 10 3 kg and a charge of q = 1 . 22 10 8 C has, at a given instant, a velocity...

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