A charged particle is moving in a, uniform, constant magnetic field. Which one o
ID: 1997625 • Letter: A
Question
A charged particle is moving in a, uniform, constant magnetic field. Which one of the following statements concerning the magnetic force exerted on the particle in false? It does no work on the particle. It increases the speed of the particle. It changes the velocity of the particle. It can act only on a particle in motion. It does not change the kinetic energy of the particle. A long, straight wire is in the same plane as a conducting loop. The wire carries an increasing current I in the direction shown in the figure. There will be a clockwise induced current in the loop. There will be a counterclockwise induced current in the loop. There will he no induced emf and no induced current. There will be a clockwise induced emf. but no induced current. There will be a counterclockwise induced emf. but no induced current. When an RLC series circuit is in resonance, its impedance is: pi/2 ohm - pi/2 ohm a maximum zero equal to its resistance Problems 6-15 are worth a total of 85 points. Provide your answer on the space provided. Show your work or explain your answers for partial to full credit; that includes showing the equations used. Unsupported answers could get no credit. An electron moves through a region of crossed electric and magnetic fields. The electric field E = 6000 V/m and is directed straight down. The magnetic field B = 1.5 T and is directed to the left. For what velocity v of the electron into the paper will the electric force exactly cancel the magnetic force?Explanation / Answer
3. F = q (v x B)
force will always be perpendicular to the velocity of particle.
hence magnitdue of velocity or speed will be constant hence no change in KE Or no work will done.
but direction changes hence velocity changess.
Ans. (a, c, d, e )
4. field will be into the page.
and current increases so flux is increases (into the page.)
so induced current will be like that field will be in out of page direction.
Ans. counterclockwise.
(B)
5 , Z = R
ans(d)
6 . Fb = Fe
q ( v x B) = q E
E = v B
v = E/B = 4000 m/s
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