# hw9 - Homework 9 Solutions Spring 2010 Required Problems...

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Unformatted text preview: Homework 9 Solutions Spring 2010 Required Problems Chapter 27 6. The magnetic force must be equal in magnitude to the force of gravity on the wire. The maximum magnetic force is applicable since the wire is perpendicular to the magnetic field. The mass of the wire is the density of copper times the volume of the wire. ( 29 ( 29 ( 29 ( 29 ( 29 2 1 B 2 2 3 3 3 2 2 5 8.9 10 kg m 1.00 10 m 9.80m s 1400A 4 4 5.0 10 T F mg I B d g d g I B -- = = = = = l l This answer does not seem feasible. The current is very large, and the resistive heating in the thin copper wire would probably melt it. 12. The net force on the current loop is the sum of the infinitesimal forces obtained from each current element. From the figure, we see that at each current segment, the magnetic field is perpendicular to the current. This results in a force with only radial and vertical components. By symmetry, we find that the radial force components from segments on opposite sides of the loop cancel. The net force then is purely vertical. Symmetry also shows us that each current element contributes the same magnitude of force. ( 29 ( 29 2 2 2 sin 2 2 r r Id B IB d I B r IB r d = = - = - = - + F k k k L l r r 19. ( a ) The velocity of the ion can be found using energy conservation. The electrical potential energy of the ion becomes kinetic energy as it is accelerated. Then, since the ion is moving perpendicular to the magnetic field, the magnetic force will be a maximum. That force will cause the ion to move in a circular path. 2 1 initial final 2 2 qV E E qV mv v m = = = ( 29 ( 29 ( 29 2 max 27 2 19 2 2 6.6 10 kg 2700V 1 2 1 3.1 10 m 0.340T 2 1.60 10 C v F qvB m r qV m mv mV m r qB qB B q--- = = = = = = = ( b ) The period can be found from the speed and the radius. Use the expressions for...
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## hw9 - Homework 9 Solutions Spring 2010 Required Problems...

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