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Questions
In a chamber, a uniform magnetic field of 6.5 G (1 G = 10–4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 106 m s−1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.5 × 10–19 C, me = 9.1 × 10–31 kg)
In a chamber, a uniform magnetic field of 6.5 G (1 G = 10–4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 106 m s−1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.6 × 10–19 C, me = 9.1 × 10–31 kg)
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Solution 1
Magnetic field strength, B = 6.5 G = 6.5 × 10–4 T
Speed of the electron, v = 4.8 × 106 m/s
Charge on the electron, e = 1.5 × 10–19 C
Mass of the electron, me = 9.1 × 10–31 kg
Angle between the shot electron and magnetic field, θ = 90°
Magnetic force exerted on the electron in the magnetic field is given as:
F = evB sin θ
This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.
Hence, the centripetal force exerted on the electron,
`"F"_"e" = ("mv"^2)/"r"`
In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,
`"F"_"e"= "F "`
`("mv"^2)/"r" = "evB" sintheta`
`"r" = ("mv")/("B" "e"sin theta)`
= `(9.1 xx 10^-31 xx 4.8 xx 10^6)/(6.5 xx 10^-4 xx 1.5 xx 10^-19 xx sin 90°)`
= 4.48 × 10–2
= 45 mm
Hence, the radius of the circular orbit of the electron is 45 mm.
Solution 2
Magnetic field strength, B = 6.5 G = 6.5 × 10–4 T
Speed of the electron, v = 4.8 × 106 m/s
Charge on the electron, e = 1.6 × 10–19 C
Mass of the electron, me = 9.1 × 10–31 kg
Angle between the shot electron and magnetic field, θ = 90°
Magnetic force exerted on the electron in the magnetic field is given as:
F = evB sin θ
This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.
Hence, the centripetal force exerted on the electron,
`"F"_"e" = ("mv"^2)/"r"`
In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,
`"F"_"e"= "F "`
`("mv"^2)/"r" = "evB" sintheta`
`"r" = ("mv")/("B" "e"sin theta)`
= `(9.1 xx 10^-31 xx 4.8 xx 10^6)/(6.5 xx 10^-4 xx 1.6 xx 10^-19 xx sin 90°)`
= 4.2 × 10–2 m
= 4.2 cm
Hence, the radius of the circular orbit of the electron is 4.2 cm.
Notes
(e is taken as either e = 1.5 × 10–19 C or 1.6 × 10–19 C)
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