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

the part of the wave he is travelling on is the crest

Answer from: Quest

5.5 n====== $\sum f = ma$ assuming that $\sum f = f_{app}$ we have m = 0.145 kg a = 38 m/s²therefore $f_{app} = (0.145\text{ kg})(38\text{ m/s}^2) = 5.51\text{ n}$ 5.51 n, but two significant figures.final answer: 5.5 n

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