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Question
(a) A monoenergetic electron beam with electron speed of 5.20 × 106 m s−1 is subject to a magnetic field of 1.30 × 10−4 T normal to the beam velocity. What is the a radius of the circle traced by the beam, given e/m for electron equals 1.76 × 1011 C kg−1?
(b) Is the formula you employ in (a) valid for calculating the radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
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Solution
(a) Speed of an electron, v = 5.20 × 106 m/s
Magnetic field experienced by the electron, B = 1.30 × 10−4 T
Specific charge of an electron, e/m = 1.76 × 1011 C kg−1
Where,
e = Charge on the electron = 1.6 × 10−19 C
m = Mass of the electron = 9.1 × 10−31 kg−1
The force exerted on the electron is given as:
`"F" = "e"|vec"v" xx vec"B"|`
= evB sin θ
θ = Angle between the magnetic field and the beam velocity
The magnetic field is normal to the direction of beam.
∴ θ = 90°
F = evB .............(1)
The beam traces a circular path of radius, r. It is the magnetic field, due to its bending nature, that provides the centripetal force `("F" = ("mv"^2)/"r")` for the beam.
Hence, equation (1) reduces to:
`"evB" = ("mv"^2)/"r"`
∴ r = `"mv"/("eB") = "v"/(("e"/"m")"B")`
= `(5.20 xx 10^6) /((1.76 xx 10^11) xx 1.30 xx 10^(-4))`
= 0.227 m
= 22.7 cm
Therefore, the radius of the circular path is 22.7 cm.
(b) Energy of the electron beam, E = 20 MeV = 20 × 106 × 1.6 × 10−19 J
The energy of the electron is given as:
`"E" = 1/2 "mv"^2`
∴ v = `((2"E")/"m")^(1/2)`
= `sqrt((2xx20xx10^6xx 1.6 xx 10^(-19))/(9.1 xx 10^(-31)))`
= 2.652 × 109 m/s
This result is incorrect because nothing can move faster than light. In the above formula, the expression (mv2/2) for energy can only be used in the non-relativistic limit, i.e., for v << c.
When very high speeds are concerned, the relativistic domain comes into consideration.
In the relativistic domain, mass is given as:
`"m" = "m"_0 [1 - "v"^2/"c"^2]^(1/2)`
Where,
`"m"_0` = Mass of the particle at rest
Hence, the radius of the circular path is given as:
r = `"mv"/"eB"`
= `("m"_0"v")/("eB"sqrt(("c"^2 - "v"^2)/"c"^2))`
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