A deuteron i accelerated from rest through a 10,000 volt potential difference and then moves perpendicularly to a uniform magnetic field of strength B = 1.6T. What is the radius of the resulting circular path? (For a deuteron: m=3.3x10^-27 kg, q= 1.6x10^-19C)
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This is way out of my training, but I'll take a crack at it since others haven't responded yet. But before I do, I'll explain how I'm coming at this.
Charged particles are under acceleration when experiencing an electric field gradient. Your problem states the total difference of potential but not the distance. Luckily, no matter how much distance is involved the final energy is the same -- a longer distance means a lower gradient and a lower acceleration but a longer distance for the acceleration, while a shorter distance means a higher acceleration but a shorter distance over which it operates. In the end, the energy is the same, which is why volts are just Joules per Coulomb and don't include other dimensions. From this, I'm gathering that this might be a particle gun that uses something like a Wehnelt with a hole in it that allows some particles though after being accelerated.
So the potential energy before starting on its trek to the Wehnelt is:
1. U = q•V
But this is converted from potential energy to kinetic energy by the time the particle exits the electric field potential, and classically this is then:
2. U = ½•m•v²
So setting those equal and solving for v gives:
3. v = √[ 2•q•V / m ]
Just as a double-check for classical analysis, this works out to less than 1000 km/s, which is well under the speed of light. So I think it's safe to proceed.
Okay. Now that is the velocity at exit into the uniform magnetic field. The force of a charge moving through the magnetic field will experience a force that is ever perpendicular to the charged particle's motion and the B-field so it cork-screws as it moves. If I'm guessing right, the circular path radius must be such that the centrifugal force matches with the resulting EM force from the moving charge and that works out to:
Charged particles are under acceleration when experiencing an electric field gradient. Your problem states the total difference of potential but not the distance. Luckily, no matter how much distance is involved the final energy is the same -- a longer distance means a lower gradient and a lower acceleration but a longer distance for the acceleration, while a shorter distance means a higher acceleration but a shorter distance over which it operates. In the end, the energy is the same, which is why volts are just Joules per Coulomb and don't include other dimensions. From this, I'm gathering that this might be a particle gun that uses something like a Wehnelt with a hole in it that allows some particles though after being accelerated.
So the potential energy before starting on its trek to the Wehnelt is:
1. U = q•V
But this is converted from potential energy to kinetic energy by the time the particle exits the electric field potential, and classically this is then:
2. U = ½•m•v²
So setting those equal and solving for v gives:
3. v = √[ 2•q•V / m ]
Just as a double-check for classical analysis, this works out to less than 1000 km/s, which is well under the speed of light. So I think it's safe to proceed.
Okay. Now that is the velocity at exit into the uniform magnetic field. The force of a charge moving through the magnetic field will experience a force that is ever perpendicular to the charged particle's motion and the B-field so it cork-screws as it moves. If I'm guessing right, the circular path radius must be such that the centrifugal force matches with the resulting EM force from the moving charge and that works out to:
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