Recall that the density J of current flowing through a material can be written i
ID: 2005144 • Letter: R
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Recall that the density J of current flowing through a material can be written in terms of microscopic properties of the material: j=nqvd , where n is the density of current carriers, is the charge of one current carrier, andvd is the drift velocity of a current carrier. In a metal, the current carriers are electrons.
Recall that the density J of current flowing through a material can be written in terms of microscopic properties of the material: j=nqvd , where n is the density of current carriers, is the charge of one current carrier, and vd is the drift velocity of a current carrier. In a metal, the current carriers are electrons. The drift velocity is the component of the current - carrier's velocity due to acceleration from the electric field in the conductor. This corresponds to the average speed of all of the current carriers in the conductor. The current carriers also have random thermal motions, but the randomness causes the velocities due to thermal motion to cancel when averaged over a large number of current carriers. If the electric field inside of the conductor has magnitude E, and the charge q is accelerated from rest for a time tau, what is the final speed v of the charge? Part A: Express the speed in terms of E, q, tau, and the mass m of the charge. Part B: The magnitude of the drift velocity is very small compared to the speed of random electron motion in a metal. The mean time between collisions can be calculated from the mean free path d, which is the average distance that an electron can travel before colliding with one of the metal nuclei. Using these variables, what is the mean time between collisions tau? Let E_F be the energy of the electrons. Express the mean time between collisions in terms of E_F, d, and m. tau =Explanation / Answer
Given: Charge of the particle = q since , it starts from rest , intial velocity of the particle = U = 0 m/s Electric field inside the conductor = E When the particle entres in to this electric field ,it will acting electrical force on the charged particle , F = E q But ,by the newton's second law , force F = ma where m is the mass of the particle a is the acceleration these force are equal F = ma = E q thus, acceleration attained in the particle is a = (E q / m ) let ,final velocity of the particle is = V = ? using kinematic realtion , V = U + a where for attain this acceleration thus, final velocity of the particle is V = 0 + (Eq /m) hence ,final velocity of the charged particle is V = (Eq /m ) POst B seperetely...,Related Questions
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