# hw_ch22 - Instructor Tomoyuki Nakayama Tuesday PHY 2005...

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Instructor: Tomoyuki Nakayama Tuesday, June 15, 2010 PHY 2005 Applied Physics 2 - Summer C 2010 Solutions for Suggested Homework Problems (Chapter 22) ________________________________________________________________________________ 2. The magnetic force on a wire of length L carrying current I in a uniform magnetic field B is F = IBL sin θ , where θ is the angle between the direction of the magnetic field and that of the current. The direction of the force is given by the right hand rule. a) F = IBLsin 90° = 0.360 N, into the page b) F = IBL sin50° = 0.276 N, into the page c) F = IBL sin0° = 0. Note 150 G = 0.015 T and 60 cm = 0.6 m. 5. The component of the field perpendicular to a current causes the magnetic force; this is because B sin θ = B . The force on the vertical wire is F = IB x L = 10.2 × 10 -4 N, where B x is the horizontal component of the field. The right hand rule tells you that the force is directed to the west. 10. The magnetic force on charge q moving with a speed v in a magnetic field B is F = qvB sin θ , where θ is the angle between the direction of the field and that of the velocity. a) A negative charge moving in one direction is equivalent to a positive charge with the same magnitude moving in the opposite direction. So point your thumb in the direction opposite the velocity of the electron. The magnetic force is directed out of the page. The magnitude is F = evB sin20° = 5.75 × 10 -15 N b) The right hand rule tells you that the force is directed into the page. The magnitude of the force is the same as that on the electron because they have the same amount of charge:

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## This note was uploaded on 07/16/2011 for the course PHY 2005 taught by Professor Lee during the Summer '08 term at University of Florida.

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hw_ch22 - Instructor Tomoyuki Nakayama Tuesday PHY 2005...

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