Question

Consider an electron in a cyclotron motion with the velocity v = 9.000 × 10 6 m/s v=9.000×106 m/s perpendicular to the magnetic field of strength B = 1.00 × 10 − 3 T. B=1.00×10−3 T. Find the ratio of the field created during the cyclotron motion to the external field.

Answer 1

The ratio of the field created during the cyclotron motion to the external field is; 55 × 10⁻¹⁵

We are given;

Velocity; v = 9 × 10⁶ m/s

B_ext = 1 × 10⁻³ T

Since we are dealing with a circular motion, then Force is;

F = mv²/r

Where;

m is mass of electron = 9.1 × 10⁻³¹ Kg

v is velocity

r is distance

Also since we are also dealing with a magnetic field, then force is;

F = qVB

Where;

q is charge on electron = 1.6 × 10⁻¹⁹ C

V = 9 × 10⁶ m/s

B is magnetic field strength

v is velocity

Thus;

mv²/r = qvB

Divide both sides by v to get;

mv/r = qB

r = mv/qB

plug in relevant values to get;

r = (9.1 × 10⁻³¹ × 9 × 10⁶)/(1.6 × 10⁻¹⁹ × 1 × 10⁻³)

r = 0.0512 m

To find the formula for the field created during the cyclotron motion, we will use the bio savart's formula;

B = (μ_o * I)/2r

where;

μ_o = 4π × 10⁻⁷ H/m

I is current

r is distance

We don't have current and we will get it from the equation;

I = qv/(2πr)

I = (1.6 × 10⁻¹⁹ × 9 × 10⁶)/(2π × 0.0512)

I = 4.47 × 10⁻¹² A

Thus;

B = (4π × 10⁻⁷ × 4.47 × 10⁻¹²)/(2 × 0.0512)

B = 55 × 10⁻¹⁸ T

Thus;

the ratio of the field created during the cyclotron motion to the external field = (55 × 10⁻¹⁸)/(1 × 10⁻³ )

⇒ 55 × 10⁻¹⁵

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Answer 2

The ratio of field created during the cyclotron motion to the external field. is $5.5 5\times 10^{-14}$.

The given parameters;

• speed of the electron, v = 9 x 10⁶ m/s
• magnetic field strength, B = 1.0 x 10⁻³ T

The magnetic force on the electron is calculated as follows;

F = qvB

The centripetal force on the electron is calculated as follows;

$F = \frac{mv^2}{r}$

Solve the two equations together, to get the radius of the circular path;

$\frac{mv^2}{r} = qvB\\\\r = \frac{mv^2}{qvB} \\\\r = \frac{mv}{qB} \\\\r = \frac{9.11 \times 10^{-31} \times 9\times 10^6}{1.602 \times 10^{-19} \times 1\times 10^{-3}} = 0.051 \ m$

The current flowing in the circular path can be determined using the following magnetic moment relations;

$I = \frac{qv}{2\pi r} \\\\I = \frac{1.602 \times 10^{-19} \times 9\times 10^6}{2\pi \times 0.051} \\\\I = 4.5 \times 10^{-12} \ A$

The magnetic field created in the circular loop is determined using Biot-Savart law;

$B = \frac{\mu_0 I}{2r} \\\\B = \frac{(4\pi \times 10^{-7} )\times (4.5 \times 10^{-12} )}{2\times 0.051} \\\\B = 5.55 \times 10^{-17} \ T$

The ratio of the field created during the cyclotron motion to the external field is calculated as;

$\frac{B_i}{B_e} = \frac{5.55 \times 10^{-17}}{1.0 \times 10^{-3}} = 5.55 \times 10^{-14}$

Thus, the ratio of field created during the cyclotron motion to the external field. is $5.5 5\times 10^{-14}$.

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