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elena55 [62]
4 years ago
7

It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, w

ould the velocity of a neutral particle be selected by passage through this device?) The explanation of this is that the mass and the charge control the resolution of the device--particles with the wrong velocity will be accelerated away from the straight line and will not pass through the exit slit. If the acceleration depends strongly on the velocity, then particles with just slightly wrong velocities will feel a substantial transverse acceleration and will not exit the selector. Because the acceleration depends on the mass and charge, these influence the sharpness (resolution) of the transmitted particles.
Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with __________.

a) both q and m large
b) q large and m small
c) q small and m large
d) both q and m small
Physics
1 answer:
Charra [1.4K]4 years ago
6 0

Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

q > m

As we know the formula:

F = Eq

And we also know that :

F = Bqv

F = \frac{mv^{2} }{r}

Bqv = \frac{mv^{2} }{r}

or Eq = \frac{mv^{2} }{r}

Assume that you want a velocity selector that will allow particles of velocity v⃗  to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

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Answer:

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N= number of turns

l = length of the solenoid

d= diameter of the solenoid

A=cross section area

B=magnetic induction

\phi = magnetic flux

I= Current

Given that, Solenoid A has total number of turns N, length L and diameter D

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Solenoid B has total number of turns 2N, length  2L and diameter 2D

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\frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=\frac{\mu_0\pi\frac{N^2D^2}{4L}}{\mu_0\pi\frac{16 N^2D^2}{4.2L}}

\Rightarrow \frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=\frac18

\Rightarrow  {\textrm{Inductance of A}}=\frac18\times {\textrm{Inductance of B}}

∴Inductance of solenoid A is \frac18 of inductance of solenoid B.

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Answer:d

Explanation:

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