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course Phy 2012
Hello Dr. Smith, The following problems I am not having trouble with. Could you please help me? I am taking this exam on Thursday .Thanks so much.
"A proton (mass 1.67 * 10^-27 kg) passes to the East through a magnetic field of .0075 Tesla directed vertically downward, crossed with an electric field directed either North or South. The proton passes through undeflected at a velocity of 4.3 * 10^8 m/s. Is the electric field directed North or South, and what is the field strength? What force does the proton experience from each field?
I am not 100% how to proceed with this problem.
q v B = q E
v B= E
(4.3*10^8 m/s)* (0.0075 T)= 3225000 N/c
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You have the electric field.
You don't have the reasoning for the direction of the electric field.
The force exerted by the magnetic field is in the direction of the vector q `v X `B, which is perpendicular to both q `v and `B by the right-hand rule. The direction of q `v is the same as the direction of `v, since q is positive for a proton. Thus q `v is toward the east. The magnetic field is directed downward, so by the right-hand rule the force is to the north.
The force exerted by the electric field is in the direction of the field, since the proton has a positive charge. The force exerted by the electric field is equal and opposite to that exerted by the magnetic field, since the proton passes through undeflected.
It follows that the electric field is directed to the south.
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I am not sure what else to do for this problem, could you guide me?
Charges of -2.251 `microCoulombs and 2.25 `microCoulombs are located at x = 0 and x = 22 meters, respectively. The charge at x = 0 is fixed at that position.
• What is the force exerted on the 2.25 `microCoulomb charge by the charge at x = 0?
• What is the electric field at the 22 meter position, in Newtons / Coulomb?
• If the 2.25 `microCoulomb charge resides on an object of mass .0038 kg, then if that charge is released approximately how fast will it be moving after traveling 1 cm, assuming that each charge experiences only the forces exerted on it by the other?
I am not sure how to do this problem???
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This is an Introductory Problem Set problem, so be sure you can do all the problems in that set, and that you understand the solutions to all those problems.
This problem uses Coulomb's Law to find the force.
The electric field at a point, being the force per unit charge, is equal to the force on a charge at that point, divided by the charge.
The work done on the charge by the force as is moves 1 cm is the product of the force and the distance. This product will be equal to the kinetic energy of the particle, by the work-kinetic energy theorem.
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A hypothetical atom with negligible kinetic energy has a mass of 212 amu. It undergoes a gamma decay. The remaining atom has atomic mass which is less than that of the original by .0000146 amu. What is the kinetic energy and/or wavelength (whichever is more appropriate) of the emitted particle, assuming that the kinetic energy of the remaining atom is negligible? How much energy would be released by the decay of a mole of these atoms? Note that the mass of a helium nucleus is about 4.001 amu and the mass of an electron about .00055 amu, where an amu is approximately 1.66 * 10^-27 kg?
Also here I am not sure what to do???
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This is an introductory problem set problem, but it's for the next test and shouldn't appear on this test. If a problem of this nature does appear, it will be disregarded (unless you happen to get it right, in which case I'll cerntainly count it).
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An electron (mass 9.11 * 10^-31 kg) with velocity directed to the North passes through a magnetic field of .0019 Tesla directed vertically upward, crossed with an electric field of 63000 N/C directed either East or West. The electron passes through undeflected. Is the electric field directed East or West, and how fast is the electron moving? What force does the electron experience from each field?
v=E/B
v=63000N/C*0.0019T
v=
I am not sure how to tell the direction or if I am in the right direction, that certainly confuses me?!"
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If you reduce Newtons and Tesla to fundamental units you will find that N / C divided by T gives you m/s. So you will have the correct velocity.
You need to use the right-hand rule to readon this out. The reasoning is similar to that of the earlier problem, but note that direction of the vector q `v is opposite to that of `v, since the charge q of the electron is negative.
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Check my notes, and be sure to really know the introductory problem set.
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