Electron Motion in a Magnetic Field: Kinetic Energy Considerations

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In a magnetic field oriented in the positive x-direction, an electron moving in the positive y-direction experiences a magnetic force that causes it to move into the paper. According to Fleming's left-hand rule, the first finger indicates the magnetic field, the second finger represents the current (electron's velocity), and the thumb shows the direction of the force acting on the electron. This results in a circular motion of the electron within the magnetic field. The kinetic energy of the electron remains constant as it moves, since the magnetic force does not do work on the electron. The discussion emphasizes the relationship between magnetic forces and electron motion, highlighting the principles of electromagnetism.
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Consider a region in which there is a magnetic field B, which is such that it has magnitude B and direction in the positive x-direction. An electron enters this region such that its velocity is in the positive y-direction. (a) Describe the motion of the electron in this region by considering the magnetic force on it. (Hint: as a third direction use into the paper or out of the paper). (b) What can you say about the kinetic energy of the electron?
 
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You have to use flemings left hand rule:

http://en.wikipedia.org/wiki/Fleming's_left_hand_rule

so your 1st finger (field) points in positive x direction (left to right on paper)

your second finger (current) points upwards towards the top of the paper

and your thumb (motion) points INTO the paper.

just be careful with this one, you can seriously sprain your fingers! :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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