Electricity and Magnetism homework

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Homework Statement


What is the Kinetic Energy of an electron that passes undeviated through perpendicular electric and magnetic fields if E = 4.0 kV/m and B = 8.0 mT?
[tex]m_{e} = 9.1 \times 10^{-31} \textup{kg}[/tex]
[tex]q_{e} = 1.6 \times 10^{-19} \textup{C}[/tex]


Homework Equations


I got stuck doing this problem so can someone please help me solve it?


The Attempt at a Solution


First what I did was I converted the 4.0 kilovolts to 4000 volts and changed to equivalent units - N/C. Then, I used the equation - [tex]\textup{E}= \frac{F}{q}[/tex] to figure out the force on the charge from the electrical field. I found that to be [tex]6.4 \times 10^{-16} \textup{N}[/tex]... Then, I tried finding the force from the magnetic field and got stuck because the equation - [tex]B = \frac{F}{qv}[/tex] requires velocity... The velocity wasn't given and I don't know how to figure it out...

Homework Statement


An electron moves through a region of crossed electric and magnetic fields. The electric field E = 1000 V/m and is directed straight down. The magnetic field B = 0.4 T and is directed to the left. For what velocity V of the electron into the paper will the electric force exactly cancel the magnetic force?

mag.jpg


Homework Equations


Please help me solve this problem because I have no idea where to start.


The Attempt at a Solution


I don't have an attempt because I don't know where to start solving the problem.
 
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Answers and Replies

  • #2
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Picture an electron moving to the right, the magnetic field is pointing down the page, therefore the magnetic force is out of the page and is equal to qvB. Then to cancel this, the electric field must also be out of the page to produce a force into the page equal to qE. Then we have qvB = qE, then what must v be equal to? And then what must K.E. be equal to?

This reasoning will solve both problems.
 
  • #3
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Picture an electron moving to the right, the magnetic field is pointing down the page, therefore the magnetic force is out of the page and is equal to qvB. Then to cancel this, the electric field must also be out of the page to produce a force into the page equal to qE. Then we have qvB = qE, then what must v be equal to? And then what must K.E. be equal to?

This reasoning will solve both problems.
thanks for the reply...
i just tried solving the two problems and qvb=qE so then I set the first problem up like this:

[tex]1.6\times 10^{-19 } \textup{ C }\times V \times 8^{-3}\textup{ T} = 1.6\times 10^{-19} \textup{ C}\times 4000\textup{ V/m}[/tex]
I also have a question about this... Don't we have to add the forces on the electron from the electrical field and the magnetic field? In other words, [tex]F = F_{electrical} + F_{magnetic}[/tex]
 
  • #4
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thanks for the reply...
i just tried solving the two problems and qvb=qE so then I set the first problem up like this:

[tex]1.6\times 10^{-19 } \textup{ C }\times V \times 8^{-3}\textup{ T} = 1.6\times 10^{-19} \textup{ C}\times 4000\textup{ V/m}[/tex]
I also have a question about this... Don't we have to add the forces on the electron from the electrical field and the magnetic field? In other words, [tex]F = F_{electrical} + F_{magnetic}[/tex]
This is exactly what we are doing

[tex] \vec F_{net} = \vec 0 = \vec F_E + \vec F_B = q \vec E - q \vec v \times \vec B [/tex]

then, since the forces are both lie along the same line, we evaluate the magnitudes of the forces as

[tex] qE - qvB = 0 [/tex]
 
  • #5
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thanks for the explanation...
I also have two other problems that I'm really not sure about. If you could explain them to me I would really appreciate it.

Homework Statement


Two long parallel wires separated by 4.0 mm each carry a current of 24 Amps. These two currents are in the same direction. What is the magnitude of the magnetic field at a point that is between the two wires and 1.0 mm from one of the two wires? Note - [tex]\mu _{0}[/tex] = permeability of free space and is defined to have the value of
[tex]4\pi \times 10^{-7} T\cdot m/A[/tex]

Homework Equations


Magnetic field between two parallel conductors - [tex]B = \frac{\mu _{0}I}{2\pi d}[/tex]
Magnetic field at a distance r from 1 wire carrying a current I is [tex]B = \frac{\mu _{0}I}{2\pi r}[/tex]

The Attempt at a Solution


My attempt at solving this problem began with using the first equation above - [tex]B = \frac{\mu _{0}I}{2\pi d}[/tex] : [tex]B = \frac{4\pi \times 10^{-7} \times 24 \textup{ Amps}}{2\pi \times 4^{-3}\textup{ m}}[/tex] and here, because I used [tex]4^{-3}\textup{ m}[/tex] for 'd' in the equation I believe I found the density of the magnetic field between the wires... However, when instead of the [tex]4^{-3}\textup{ m}[/tex] I used the [tex]1^{-3}\textup{ m}[/tex] it represents r or the magnetic field a distance away from only 1 of the wires?

Homework Statement


Two long parallel wires each carry a current of 5.0 Amps. These two currents are oppositely directed. The two wires are separated by 6.0 cm. What is the magnitude of the magnetic field at a point that is 5.0 cm from each of the wires?

Homework Equations


For this problem I used the same equations above.

The Attempt at a Solution


I'm not sure how to solve this problem because first the currents are in opposite directions which would make one I be positive and the other I negative? Then, I'm not sure what to do when in the question it asks 5.0 cm from each of the wires? Does the word 'each' signify that I have to add the magnetic field density 5.0 cm from each wire together?
 
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  • #6
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so can someone please help me with the two problems?
 

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