My answers so far are:
a) J=I/(2*pi*r*t)
b) u=J/ne=I/(2*pi*r*t*n*e)
c) F=IB/(2*pi*r*t*n), in theta direction (polar coordinates)
d) This is where I am stuck.
I understand the example for motional emf with a rod moving through a magnetic field but I'm not sure how to apply it to this scenario...
liquid
melting point (degrees C)
boiling point (degrees C)
water (H2O)
0
100
sodium (Na)
98
883
Sodium-potassium(NaK)
-11
785
Lead(Pb)
327
1749
I'm prettttty sure by consulting the literature means by using the above table… but if that's the case then how in the world do you find Cv...
oh yeahhh of course, that was stupid of me, I found the particular solution ## v_{xp}=\frac{V}{bd} ##, but forgot to include it.
Awesome, I get the same solutions for vx, vy and s as you and delta^2 too now.
That being vx=-V/(Bd)cos(Bqt/m)+V/Bd and the other two as stated.
Thanks heaps again...
ok so i got ## v_x ## = Acos(qbt/m)+Csin(qbt/m) for the homogeneous solution, but if ## v_x ## = 0, when t=0, then doesn't that imply A=0 so we eliminate the Acos(qbt/m) from our equation instead of the sin? although i think this is wrong because this leads me to get ## v_x ## = V/Bd and that...
oh yep, i forgot that minus sign, i wondered why i had an extra one later one.
what you've said in your second post is what i said before but maybe i didn't make it clear.
kinetic energy increase is the decrease in potential energy from the electric field which is qV, and so we have...
Ok…so i subbed in ##\vec{v}=v_x\vec{i}+v_y\vec{j}## and got $$m_e\frac{d\v_x\vec{i}}{dt}=q\frac{V}{d}\vec{j}+q\vec{v}\times \vec{B} $$ and $$m_e\frac{d\v_y\vec{j}}{dt}=q\vec{j}(frac{V}{d}+\vec{v_y}\times \vec{B}) $$ and said that for the electron to just not touch the top plate, the force in the...
Well initially it is to the right, but as the electron changes direction the magnetic force direction will also change, swinging around until it points down as the electron just nearly touches the plate, it means the radius will be curved but it won't be a circular trajectory because of the...
Homework Statement
In the figure, an electron of mass m, charge − e, and low (negligible) speed enters the region between two plates of potential difference V and plate separation d, initially headed directly toward the top plate. A uniform magnetic field of magnitude B is normal to the plane...