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# OHW 17 - terry(ect328 oldhomework 17 Turner(59130 This...

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terry (ect328) – oldhomework 17 – Turner – (59130) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An electron is in a uniform magnetic field B that is directed into the plane of the page, as shown. v e - B B B B When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed 1. toward the right 2. toward the top of the page. 3. into the page. 4. toward the bottom of the page. correct 5. out of the page. 6. toward the left Explanation: The force on the electron is F = q v × B = - e v × B. The direction of the force is thus F = - v × B , pointing toward the bottom of the page , us- ing right hand rule for v × B , and reversing the direction due to the negative charge on the electron. 002 10.0 points A particle with charge q and mass m is un- dergoing circular motion with speed v . At t = 0, the particle is moving along the nega- tive x axis in the plane perpendicular to the magnetic field B , which points in the positive z direction in the figure below. x y z v B Find the direction of the instantaneous ac- celeration a at t = 0 if q is negative. 1. a = ˆ k 2. a = - ˆ k 3. a = ˆ k + ˆ i 4. a = - ˆ i 5. a = ˆ j 6. a = ˆ i 7. a = - ˆ k + ˆ i 8. a = ˆ i + ˆ j 9. a = - ˆ j correct 10. a = ˆ j + ˆ k Explanation: The particle is moving along the negative x -axis in this instant v = - v ˆ i ; since it is moving in a circle, we need to talk about instantaneous direction. The force F B is equal to q v × B at all times. We know that B is pointing in the z direction, so B = B ˆ k , and therefore F B = q v ( - ˆ i ) × B ˆ k

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terry (ect328) – oldhomework 17 – Turner – (59130) 2 = q v B ( - ˆ i × ˆ k ) = q v B ˆ j .
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OHW 17 - terry(ect328 oldhomework 17 Turner(59130 This...

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