Answer :
Explanation:
This problem can be solved by using the formula
[tex]r = \frac{mv}{qB}[/tex]
where m is the mass of the ion, v is the velocity, q is the charge, B is the strength of the magnetic field and r is the radius of the circular motion of the ion.
(a) Taking q from the equation and by replacing we have
[tex]q = \frac{mv}{rB} = \frac{(2.66*10^{-26})(5*10^{6})}{(0.231)(1.20)} = 4.79*10^{-9} C[/tex].
(b) [tex]\frac{4.79*10^{-9}}{q_{e}} = \frac{4.79*10^{-9}}{1.6*10^{-19}} } = 2.99*10^{10}[/tex] ≈ 3*10^(10) C
the ion has 2.99*10^(10) more charge than an electron .
(c) The ratio must be an integer because the electrons are indivisible.
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(a) The ratio of this charge to the charge of electron is [tex]4.8 \times 10^{-19} \ C[/tex].
(b) The ratio of this charge to the charge of electron is 3.
(c) The ratio must be integer because ion of oxygen is a multiple of charge of electron.
The given parameters;
- mass of the oxygen ion, = 2.66 x 10⁻²⁶ kg
- velocity of the ion, v = 5 x 10⁶ m/s
- magnetic field, B = 1.2 T
- radius of the arc, r = 0.231 m
The charge of the ion is calculated as follows;
[tex]F_c = F_B\\\\ \frac{mv^2}{r} = qvB\\\\q = \frac{mv}{rB} \\\\q = \frac{(2.66\times 10^{-26}) \times 5\times 10^6}{0.231 \times 1.2} \\\\q = 4.8 \times 10^{-19} \ C[/tex]
The ratio of this charge to the charge of electron is calculated as;
[tex]ratio = \frac{4.8 \times 10^{-19}}{1.6 \times 10^{-19} } = 3[/tex]
The ratio must be integer because ion of oxygen is a multiple of charge of electron.
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