Electron Force B-Field: Direction & Velocity

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The discussion centers on the direction of the force acting on an electron in a magnetic field. With the B-field directed into the page and the electron's velocity to the left, the initial assumption of the force being downwards is corrected. Using the right-hand rule and accounting for the electron's negative charge, the force is determined to point upwards instead. Participants clarify the application of the right-hand rule and the implications of the electron's charge on the force direction. Ultimately, the force acting on the electron is upward, contrary to initial beliefs.
UrbanXrisis
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B-Field is going into the page. An electron has a velocity to the left. Force is then downwards correct?
 
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No. The force is given by F = qV x B. Using the right-hand rule, and noting that the electron has a negative charge, I get that the force is up the page.
 
I forgot which hand is my right one. It's not the left one, right?

cookiemonster
 
My hands are faced down...B field into the page, means fingers curled, and if the thumb is pointing to the left, the the force is acting down.
 
UrbanXrisis said:
My hands are faced down...B field into the page, means fingers curled, and if the thumb is pointing to the left, the the force is acting down.

When you cross the velocity vector (pointing to the left) by the magnetic field vector (pointing into the page), your thumb ends up pointing down. But, as DocAl said, because we are looking at an electron, the value of the charge (q) is negative (q = -1.6*10^-19 C). The force is therefore pointing in the negative down direction or, as we say where I'm from, up.
 
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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