The Electron's Frequency of Revolution

Q. In a chamber, a uniform magnetic field of 6.5 G is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m/s normal to the field, and goes round a circular orbit. Obtain the frequency of revolution of the electron in its circular orbit, in MHz? (mₑ = 9.1 × 10⁻³¹ kg) expects 18 MHz

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Q. In a chamber, a uniform magnetic field of 6.5 G $B = 6.5$ G is maintained. An electron is shot $q = e$ into the field with a speed of 4.8 × 10⁶ m/s $v = 4.8 \times 10^{6}$ m/s normal to the field, and goes round a circular orbit. Obtain the frequency of revolution of the electron in its circular orbit, in MHz? (mₑ = 9.1 × 10⁻³¹ kg $m = 9.1 \times 10^{−31}$ kg)

The playground

Inputs

field

B

6.5 G

speed

v

4800000.0 m/s

mass

m

9.1e-31 kg

Outputs

frequency

ν

The solution

Solved with MagneticForce

Given

$B = 6.5$ G
$v = 4.8 \times 10^{6}$ m/s
$m = 9.1 \times 10^{−31}$ kg
$q = e$ from “An electron is shot”

Find $ν$ .

Working

$B = 6.5$ G $= 6.5 \times 10^{−4}$ T
An electron is shot $q = e$
$q = 1.60217663 \times 10^{−19}$
$q = 1.60217663 \times 10^{−19}$ C
a charge circles in a field(electron) $r = \frac{m \cdot v}{q \cdot B}$
$r = \frac{9.1 \times 10^{−31} \cdot 4.8 \times 10^{6}}{1.60217663 \times 10^{−19} \cdot 6.5 \times 10^{−4}}$
$r = 0.041942941$ m
once round the circle(electron) $v = 2 \cdot π \cdot r \cdot ν$
$4.8 \times 10^{6} = 2 \cdot 3.14159265 \cdot 0.041942941 \cdot ν$
$ν = 18.21$ MHz
$ν = 1.82138808 \times 10^{7}$ Hz $= 18.2138808$ MHz
$ν \approx 18$ MHzto 2 significant figures

Conclusion

A faster electron circles a wider orbit in the same time, so the frequency does not depend on the speed, only on the field and the electron's charge and mass.

The Ruby

1require_relative "../physics"23question :the_electron_s_frequency_of_revolution do4  kind :numerical5  describes "In a chamber, a uniform magnetic field of %{field} is maintained. An electron is shot into the field " \6            "with a speed of %{speed} normal to the field, and goes round a circular orbit. %{frequency}, in MHz? " \7            "(mₑ = %{mass})"8  given field: "6.5 G"9  given speed: "4.8e6 m/s"10  given mass: "9.1e-31 kg"11  asks frequency: "Obtain the frequency of revolution of the electron in its circular orbit"12end

1require_relative "../questions/the_electron_s_frequency_of_revolution"2require_relative "../science/moving_charges_and_magnetism/magnetic_force"34solve :the_electron_s_frequency_of_revolution, using: MagneticForce do5  objects :electron67  take field: B8  take speed: electron.speed9  take mass: electron.mass1011  deduce_from "An electron is shot", electron.charge => e1213  frequency = electron.frequency_of_revolution14  answer frequency:, in: "MHz", figures: 21516  conclusion "A faster electron circles a wider orbit in the same time, so the frequency does not depend on the " \17             "speed, only on the field and the electron's charge and mass."18end

1require_relative "../solutions/the_electron_s_frequency_of_revolution"23play_with :the_electron_s_frequency_of_revolution do4  title "The Electron's Frequency of Revolution"5end

The laws

quantity	written	units	within	from
pi	$π$	—	0 … ∞	MagneticForce
elementary charge	$e$	C	—	MagneticForce
gravity	$g$	m/s²	—	MagneticForce
magnetic field	$B$	T	—	MagneticForce
mass	$m$	kg	0 … ∞	MagneticForce
charge	$q$	C	—	MagneticForce
speed	$v$	m/s	—	MagneticForce
current	$I$	A	—	MagneticForce
length in the field	$l$	m	—	MagneticForce
force on a wire	$F$	N	—	MagneticForce
force per length	$f$	N/m	—	MagneticForce
radius of the path	$r$	m	—	MagneticForce
angle to the field	$θ$	—	0 … 3.142	MagneticForce
frequency of revolution	$ν$	Hz	—	MagneticForce
force on a wire on ground	$F_{g}$	N	—
current of ground	$I_{g}$	A	—
length in the field of ground	$l_{g}$	m	—
force on a wire on electron	$F$	N	—
current of electron	$I$	A	—
length in the field of electron	$l$	m	—
force per length on ground	$f_{g}$	N/m	—
angle to the field of ground	$θ_{g}$	—	0 … 3.142
force per length on electron	$f$	N/m	—
angle to the field of electron	$θ$	—	0 … 3.142
radius of the path of ground	$r_{g}$	m	—
mass of ground	$m_{g}$	kg	0 … ∞
speed of ground	$v_{g}$	m/s	—
charge of ground	$q_{g}$	C	—
radius of the path of electron	$r$	m	—
mass of electron	$m$	kg	0 … ∞
speed of electron	$v$	m/s	—
charge of electron	$q$	C	—
frequency of revolution of ground	$ν_{g}$	Hz	—
frequency of revolution of electron	$ν$	Hz	—

F_{w} = I_{w} \cdot l_{w} \cdot B

force on a wire across a field

f_{w} = I_{w} \cdot B \cdot \sin θ_{w}

force on each metre of wire

r_{c} = \frac{m_{c} \cdot v_{c}}{q_{c} \cdot B}

a charge circles in a field

v_{c} = 2 \cdot π \cdot r_{c} \cdot ν_{c}

once round the circle

F_{g} = I_{g} \cdot l_{g} \cdot B

force on a wire across a field(ground)

F = I \cdot l \cdot B

force on a wire across a field(electron)

f_{g} = I_{g} \cdot B \cdot \sin θ_{g}

force on each metre of wire(ground)

f = I \cdot B \cdot \sin θ

force on each metre of wire(electron)

r_{g} = \frac{m_{g} \cdot v_{g}}{q_{g} \cdot B}

a charge circles in a field(ground)

r = \frac{m \cdot v}{q \cdot B}

a charge circles in a field(electron)

v_{g} = 2 \cdot π \cdot r_{g} \cdot ν_{g}

once round the circle(ground)

v = 2 \cdot π \cdot r \cdot ν

once round the circle(electron)

q = e

An electron is shot