An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and 0.1250 T, re

Question

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

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Diễm Thu 3 years 2021-08-25T06:16:16+00:00 1 Answers 51 views 0

Answers ( )

  1. Tryphena
    0
    2021-08-25T06:17:49+00:00
    Reply

    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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