.45 m moves at a velocity = 3.50 x 10(i+j) in a uniform magnetic field B = 5.00 mT (3). (a) Find the direction and magnitude of the force experienced by this charge. (b) What direction and magnitude of electric field is needed to cancel the magnetic force on the particle?

A proton precesses with a frequency p in the presence of a magnetic field. If the intensity of the magnetic field is doubled, what happens to the precessional frequency?

Describe the following physical occurrences as events, that is, in the form (x, y, z, t): (a) A postman rings a doorbell of a house precisely at noon. (b) At the same lime as the doorbell is lung, a slice of bread pops out of a toaster that is located 10 1T1 from the door in the east direction from the door. (c) Tell seconds later, an airplane arrives at the airport, which is 10 km from the door in the east direction and 2 km to the south.

The creation and study of new and very massive elementary particles is an important part of contemporary physics. To create a particle of mass M requires an energy Mc2 . With enough energy, an exotic particle can be created by allowing a fast-moving proton to collide with a similar target particle. Consider a perfectly inelastic collision between two protons: an incident proton with mass kinetic energy K, and momentum magnitude p joins with an originally stationary target proton to form a single product particle of mass M. Not all the kinetic energy of the incoming proton is available to create the product particle because conservation of momentum requires that the system as a whole still must have some kinetic energy after the collision. Therefore, only a fraction of the energy of the incident particle is available to create a new particle. (a) Show that the energy available to create a product particle is given by Mc2=2mpc21+K2mpc2 This result shows that when the kinetic energy K of the incident proton is large compared with its rest energy mpc2, 2then M approaches (2mpK)1/2/c. Therefore, if the energy of the incoming proton is increased by a factor of 9, the mass you can create increases only by a factor of 3, not by a factor of 9 as would be expected. (b) This problem can be alleviated by using colliding beams as is the case in most modern accelerators. Here the total momentum of a pair of interacting particles can be zero. The center of mass can be at rest after the collision, so, in principle, all the initial kinetic energy can be used for particle creation. Show that Mc2=2mc2(1+Kmc2) where K is the kinetic energy of each of the two identical colliding particles. Here, if k mc2, we have M directly proportional to K as we would desire.

(a) Beta decay is nuclear decay in which an electron is emitted. If the electron is given 0.750 MeV of kinetic energy, what is its velocity? (b) Comment on how the high velocity is consistent with the kinetic energy as it compares to the rest mass energy of the electron.

Check Your Understanding What is if v=0.650c?

An electron in a TV CRT moves with a speed of 6.0107 m/s, in a direction perpendicular to Earth's field, which has a strength of 5.0105 T. (a) What strength electric field must be applied perpendicular to the Earth’s field to make the election moves in a straight line? (b) If this is done between plates separated by 1.00 cm, what is the voltage applied? (Note that TVs are usually surrounded by a ferromagnetic material to shield against external magnetic fields and avoid the need for such a collection,)

(a) Calculate the relativistic quantity =11v2/c2for 1.00-TeV protons produced at Fermilab. (b) If such a proton created a +having the same speed, how long would its life be in the laboratory? (c) How far could it travel in this time?

A chain of nuclear reactions in the Suns core converts four protons into a helium nucleus. (a) What is the mass difference between four protons and a helium nucleus? (b) How much energy in MeV is released during the conversion of four protons into a helium nucleus?

As measured by observers in a reference frame S, a particle having charge q moves with velocity v in a magnetic field B and an electric field E. The resulting force on the particle is then measured to be F = q(E + v × B). Another observer moves along with the charged particle and measures its charge to be q also but measures the electric field to be E′. If both observers are to measure the same force, F, show that E′ = E + v × B.