Physics, 27.12.2019 13:31 idk1561

An unknown particle moves in a straight line through crossed electric and magnetic fields with e = 1.5 kv/m and b = 0.034 t. if the electric field is turned off, the particle moves in a circular path of radius r = 2.7 cm. what might the particle be

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Answer from: robtodd4614

The mass of the particle depends on the type of particle, and the radius of

the path of the particle depend on its mass.

The particle might be an alpha particle.

Reasons:

The given parameters are;

Electric potential of the field, E = 1.5 kV/m

Magnetic field strength, B = 0.034 T

Radius of the circular path, r = 2.7 cm

Required:

Finding the type of particle in the question

Solution:

When moving in a straight line, we have;

B·q·v = F = q·E

Therefore;

E = B·v

When the particle moves in a circular path, we have;

$q\cdot E = B \cdot q \cdot v = \mathbf{\dfrac{m \cdot v^2}{r}}$

Therefore;

$v = \dfrac{B \cdot q \cdot r }{m}$

Which gives;

$E = \dfrac{B \cdot q \cdot r }{m} \times B = \dfrac{B^2 \cdot q \cdot r }{m}$

$\mathrm{The \ mass \ of \ the \ particle, \, m} = \mathbf{\dfrac{B^2 \cdot q \cdot r }{E}}$

q = n × e

Therefore;

$m = \mathbf{\dfrac{B^2 \cdot n \cdot e \cdot r }{E}}$

Where;

n = The number of electron charge

Which gives the mass in kilograms as follows;

$m = \dfrac{0.034^2 \times n\times 1.6 \times 10^{-19} \times 0.027 }{1500} = \mathbf{3.32928 \times 10^{-27} \times n}$

The particle is larger than a subatomic particle

Therefore;

$\dfrac{m}{n} = 3.32928 \times 10^{-27}$

For an alpha particle, we have;

m ≈ 6.645 × 10⁻²⁷ kg

n = 2

Therefore;

$\dfrac{6.645 \times 10^{-27}}{2} \approx \mathbf{3.3225 \times 10^{-27}} \approx 3.32928 \times 10^{-27}$

Therefore, the particle might be an alpha particle.

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Answer from: julio7264

Alpha particle

Explanation:

Initially, the particle moves in a straight line. This means that the electric force and the magnetic force are equal:

qE=qvB

where q is the charge, E is the electric field, v is the speed and B is the magnetic field.

In this problem,

E = 1.5 kV/m = 1500 V/m

B = 0.034 T

Solving the equation for v we find the speed of the particle

$v=\frac{E}{B}=\frac{1500 V/m}{0.034 T}=4.41\cdot 10^4 m/s$

Now the electric field is turned off, so the particle starts moving in a circular motion, where the magnetic force acts as centripetal force:

$qvB=m\frac{v^2}{r}$

where

r = 2.7 cm = 0.027 m is the radius of the circular path

Solving the problem for q/m, we find charge-to-mass ratio of the particle

$\frac{q}{m}=\frac{v}{Br}=\frac{4.41\cdot 10^4 m/s}{(0.034 T)(0.027 m)}=4.8\cdot 10^7 C/kg$

And this corresponds to the q/m ratio of an alpha particle, which has:

$q=2e=3.2\cdot 10^{-19}C\\m=4a.m.u.=6.64\cdot 10^{-27} kg$

Answer from: Quest

What are the choices?

Answer from: Quest

Is a vector quantity that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.

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An unknown particle moves in a straight line through crossed electric and magnetic fields with e = 1...