PHYS 1B Lecture Notes - Lecture 16: Right-Hand Rule

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5 Mar 2017

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Magnetic force law: experiments show: magnetic fields (produced by magnets/electric currents) exert magnetic force on a moving charged particle given by: fb = qv x b. Q is the charge of the particle. *note: one tesla is pretty large magnetic field that requires large magnets to produce, but it is similar to the field used in mri machines. Fb is in the direction of your thumb. Clicker question: suppose v for a positive charge is in the positive x direction and b is in the positive y direction. What is the direction of the force: d) in the positive z direction. Clicker question: suppose v for a positive charge is in the positive z direction and b is in the positive y direction. Clicker question: suppose v for a positive charge is in the negative y direction and b is in the positive z direction. What direction is the force in: b) the negative x direction.

Related Questions

An electron moving at a constant velocity, v = v0xhat, enters aregion in space where a magnetic field is present. The magneticfield, B, is constant and points in the z-direction. What is themagnitude and direction of the magnetic force acting on theelectron? (Use the following as necessary: v0, B for the magnitudeof the electric field, q for the charge of the electron, and m forthe mass of the electron.)
FB =

If the width of the region where the magnetic field is present isd, what is the minimum velocity the electron must have in order toescape this region? (Give the magnitude only. Use the following asnecessary: v0, B for the magnitude of the electric field, d, q forthe charge of the electron, and m for the mass of theelectron.)

vmin =

As measured by observers in a reference frame S, a particle havingcharge q moves
with velocity v in a magnetic field B and an electric field E . Theresulting force on
the particle is then measured to be F = q(E + v Ã B) . Anotherobserver moves along
with the charged particle and measures its charge to be q also butmeasures the
electric field to be E ' . If both observers are to measure thesame force, F , show that
E'=E+vÃB.

A velocity selector has an electric field of magnitude 2670 N/C, directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of 6.00 x 103 m/s, enter the velocity selector and are able to pass completely through without being deflected. When a different particle with an electric charge of +4.10 x 10-12 C enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is 1.65 x 10-9 N, pointing directly upward. What is the speed of this particle?

PHYS 1B Lecture Notes - Lecture 16: Right-Hand Rule

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