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padilas [110]
3 years ago
9

An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and

0.1250 T, respectively. The particle passes out of the electric field, but the magnetic field continues, and the particle makes a semicircle of diameter 25.05 cm.
Part A. What is the particle's charge-to-mass ratio?
Part B. Can you identify the particle?
a. can't identify
b. proton
c. electron
d. neutron
Physics
1 answer:
Kobotan [32]3 years ago
8 0

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic  fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is F_{ma}=BqV---(i)

The given formula to find the electric force is F_{el}=qE---(ii)

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

Bqv=qE\\\\v=\frac{E}{B}

v=\frac{187500}{0.125} \\\\v=15\times10^5m/s

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

F_{ce}=\frac{mv^2}{r}---(iii)

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br}  \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg

Therefore,  the particle's charge-to-mass ratio is 958.1\times10^5C/kg

b)

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

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{:\implies \quad \sf \lambda =\dfrac{c}{\omega}=\dfrac{3\times 10^{8}}{4\pi \times 10^{7}}}

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{:\implies \quad \boxed{\bf{k\approx 2.63\:m^{-1}}}}

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{:\implies \quad \boxed{\bf{E=50\sin \bigg(4\pi \times 10^{7}t-\dfrac{263x}{100}\bigg)}}}

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{:\implies \quad \sf B_{0}=\dfrac{E_{0}}{c}=\dfrac{50}{3\times 10^{8}}}

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Where G is gravitational constant, M is the mass of the asteroid, m is the mass of the moon and r is the distance between them

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Therefore, v can be isolated from equation 4:

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