Often when studying a school course of electromagnetism or at scientific research there is a need to establish the speed with which some elementary particle, for example, an electron or a proton moved.
Instruction
1. The following task is Let's say given: electric field with tension E and magnetic field with induction In, are excited perpendicularly each other. Perpendicular to them, evenly and rectilinearly charged particle with a charge of q and speed of v moves. It is required to determine its speed.
2. The decision is very simple. If the particle under the terms of a task moves evenly and rectilinearly, so its speed of v constant. Thus, according to the first law of Newton, sizes of forces operating on it are mutually counterbalanced, that is they are equal in the sum to zero.
3. What forces affect a particle? First, an electric component of Lorentz force which is calculated on a formula: Fel = qE. Secondly, a magnetic component of Lorentz force which is calculated on a formula: Fm = qvBSinα. As under the terms of a task the particle moves perpendicular to magnetic field, a corner α = to 90 degrees, and respectively, Sinα = 1. Then magnetic component of Lorentz force of Fm = qvB.
4. Electric and magnetic components counterbalance each other. Therefore, the sizes qE and qvB are in number equal. That is E = vB. Therefore, the speed of a particle is calculated on such formula: v = E/B. Having substituted in a formula of value E and B, you calculate required speed.
5. Or, for example, you have such task: a particle with a mass of m and a charge of q, moving with v speed, flew to the electromagnetic field. Its power lines (both electric, and magnetic) are parallel. The particle flew at an angle α to the direction of power lines and after that began to move with acceleration and. It is required to calculate with what speed it moved originally. According to the second law of Newton, acceleration of a body with a mass of m is calculated on a formula: = F/m.
6. You know the mass of a particle under the terms of a task, and F is the resulting (total) size of forces operating on it. In this case the particle is affected by the electric and magnetic leaving Lorentz forces: F = qE + qBvSinα.
7. But as power lines of fields (on a statement of the problem) are parallel, the vector of electric force is perpendicular to a vector of magnetic induction. Therefore, the total force of F is calculated on Pythagorean theorem: F = [(qE) ^2 + (qvBSinα)^2] ^1/2
8. Transforming, receive: am = q [E^2 +B^2v^2Sin^2α]^1/2. From where: v^2 = (a^2m^2 – q^2E^2) / (q^2B^2Sin^2α). After calculation and extraction of a square root, receive required size v.