Resonant motion and kinetic energy of charged particles in a time-dependent electromagnetic field

Explanation

Formulation of Equations of Motion

The force acting on a charged particle is the resultant force of the electrostatic force $q\boldsymbol{E}$qE due to the electric field and the Lorentz force $q(\boldsymbol{v} \times \boldsymbol{B})$q(v×B) due to the magnetic field. Calculating the components of the cross product, the equations of motion for each axis are obtained as the following system of differential equations:

$m \frac{dv_x}{dt} = q v_y B$mdtdvx​​=qvy​B

$m \frac{dv_y}{dt} = q E_0 \sin(\omega_c t) - q v_x B$mdtdvy​​=qE0​sin(ωc​t)−qvx​B

Here, we rearrange the equation using the cyclotron angular frequency $\omega_c = \frac{qB}{m}$ωc​=mqB​.

$\frac{dv_x}{dt} = \omega_c v_y \quad \dots \text{①}$dtdvx​​=ωc​vy​…①

$\frac{dv_y}{dt} = \frac{q E_0}{m} \sin(\omega_c t) - \omega_c v_x \quad \dots \text{②}$dtdvy​​=mqE0​​sin(ωc​t)−ωc​vx​…②

Solution of Differential Equations

Differentiate equation ② with respect to time $t$t, substitute equation ① into it, and resolve $v_x$vx​.

$\frac{d^2v_y}{dt^2} = \frac{q E_0 \omega_c}{m} \cos(\omega_c t) - \omega_c \frac{dv_x}{dt}$dt2d2vy​​=mqE0​ωc​​cos(ωc​t)−ωc​dtdvx​​

$\frac{d^2v_y}{dt^2} = \frac{q E_0 \omega_c}{m} \cos(\omega_c t) - \omega_c^2 v_y$dt2d2vy​​=mqE0​ωc​​cos(ωc​t)−ωc2​vy​

$\frac{d^2v_y}{dt^2} + \omega_c^2 v_y = \frac{q E_0 \omega_c}{m} \cos(\omega_c t) \quad \dots \text{③}$dt2d2vy​​+ωc2​vy​=mqE0​ωc​​cos(ωc​t)…③

Equation ③ is a non-homogeneous second-order linear differential equation. Since the angular frequency of the driving term on the right-hand side matches (resonates with) the natural angular frequency $\omega_c$ωc​ on the left-hand side, we assume that the particular solution is of the form $v_{y,p}(t) = \alpha t \sin(\omega_c t)$vy,p​(t)=αtsin(ωc​t).

Substitute this into the left-hand side of equation ③ and compare the coefficients $\alpha$α.

$v_{y,p}'(t) = \alpha \sin(\omega_c t) + \alpha \omega_c t \cos(\omega_c t)$vy,p′​(t)=αsin(ωc​t)+αωc​tcos(ωc​t)

$v_{y,p}''(t) = 2\alpha \omega_c \cos(\omega_c t) - \alpha \omega_c^2 t \sin(\omega_c t)$vy,p′′​(t)=2αωc​cos(ωc​t)−αωc2​tsin(ωc​t)

Substituting these,

$2\alpha \omega_c \cos(\omega_c t) = \frac{q E_0 \omega_c}{m} \cos(\omega_c t) \implies \alpha = \frac{q E_0}{2m}$2αωc​cos(ωc​t)=mqE0​ωc​​cos(ωc​t)⟹α=2mqE0​​

Therefore, the general solution can be expressed as the sum of the solutions and particular solutions of the homogeneous equation as follows:

$v_y(t) = C_1 \cos(\omega_c t) + C_2 \sin(\omega_c t) + \frac{q E_0}{2m} t \sin(\omega_c t)$vy​(t)=C1​cos(ωc​t)+C2​sin(ωc​t)+2mqE0​​tsin(ωc​t)

Application of Initial Conditions

At time $t=0$t=0, the particle is at rest, so $v_y(0) = 0$vy​(0)=0, which implies $C_1 = 0$C1​=0.

$v_y(t) = C_2 \sin(\omega_c t) + \frac{q E_0}{2m} t \sin(\omega_c t)$vy​(t)=C2​sin(ωc​t)+2mqE0​​tsin(ωc​t)

Next, substituting $t=0$t=0 and $v_x(0)=0$vx​(0)=0 into equation ②, we find that $\frac{dv_y}{dt}\Big|_{t=0} = 0$dtdvy​​​t=0​=0.

Differentiating the previously obtained $v_y(t)$vy​(t) and substituting $t=0$t=0, we get:

$v_y'(t) = C_2 \omega_c \cos(\omega_c t) + \frac{q E_0}{2m} \sin(\omega_c t) + \frac{q E_0 \omega_c}{2m} t \cos(\omega_c t)$vy′​(t)=C2​ωc​cos(ωc​t)+2mqE0​​sin(ωc​t)+2mqE0​ωc​​tcos(ωc​t)

$v_y'(0) = C_2 \omega_c = 0 \implies C_2 = 0$vy′​(0)=C2​ωc​=0⟹C2​=0

Therefore, $v_y(t)$vy​(t) and $v_x(t)$vx​(t) obtained from equation ① are as follows:

$v_y(t) = \frac{q E_0}{2m} t \sin(\omega_c t)$vy​(t)=2mqE0​​tsin(ωc​t)

$v_x(t) = \int \omega_c v_y(t) dt = \frac{q E_0 \omega_c}{2m} \int t \sin(\omega_c t) dt$vx​(t)=∫ωc​vy​(t)dt=2mqE0​ωc​​∫tsin(ωc​t)dt

Performing integration by parts and using $v_x(0) = 0$vx​(0)=0, we obtain the following equation:

$v_x(t) = \frac{q E_0}{2m} \left( \frac{1}{\omega_c} \sin(\omega_c t) - t \cos(\omega_c t) \right)$vx​(t)=2mqE0​​(ωc​1​sin(ωc​t)−tcos(ωc​t))

Calculation of Kinetic Energy

Calculate the sum of squares of velocities $v^2 = v_x^2 + v_y^2$v2=vx2​+vy2​.

$v_x^2 + v_y^2 = \left(\frac{q E_0}{2m}\right)^2 \left[ \left( \frac{1}{\omega_c} \sin(\omega_c t) - t \cos(\omega_c t) \right)^2 + t^2 \sin^2(\omega_c t) \right]$vx2​+vy2​=(2mqE0​​)2[(ωc​1​sin(ωc​t)−tcos(ωc​t))2+t2sin2(ωc​t)]

Expanding the parentheses and rearranging, we get the following relationship from the trigonometric function $\sin^2\theta + \cos^2\theta = 1$sin2θ+cos2θ=1:

$v^2 = \left(\frac{q E_0}{2m}\right)^2 \left[ \frac{1}{\omega_c^2} \sin^2(\omega_c t) - \frac{t}{\omega_c} \sin(2\omega_c t) + t^2 \right]$v2=(2mqE0​​)2[ωc2​1​sin2(ωc​t)−ωc​t​sin(2ωc​t)+t2]

The time we want to find is $t = \frac{10\pi}{\omega_c}$t=ωc​10π​, after 5 periods have elapsed. At this point, $\sin(\omega_c t) = 0$sin(ωc​t)=0 and $\sin(2\omega_c t) = 0$sin(2ωc​t)=0, so the terms are significantly simplified.

$v^2 = \left(\frac{q E_0}{2m}\right)^2 t^2 = \left(\frac{q E_0}{2m}\right)^2 \left(\frac{10\pi}{\omega_c}\right)^2 = \left(\frac{5\pi q E_0}{m \omega_c}\right)^2$v2=(2mqE0​​)2t2=(2mqE0​​)2(ωc​10π​)2=(mωc​5πqE0​​)2

Substituting $\omega_c = \frac{qB}{m}$ωc​=mqB​,

$v^2 = \left(\frac{5\pi E_0}{B}\right)^2 = \frac{25\pi^2 E_0^2}{B^2}$v2=(B5πE0​​)2=B225π2E02​​

Therefore, the kinetic energy $K$K at this time is derived as follows:

$K = \frac{1}{2} m v^2 = \frac{25 \pi^2 m E_0^2}{2B^2}$K=21​mv2=2B225π2mE02​​

Numerical Substitution

Substitute the constraint values ​​into the obtained algebraic expression to find the coefficient $A$A. $A = \frac{25 m E_0^2}{2B^2}$A=2B225mE02​​

• Numerator: $25 \times (5.00 \times 10^{-4}) \times 600^2 = 25 \times (5.00 \times 10^{-4}) \times (3.60 \times 10^5) = 25 \times 180 = 4500$25×(5.00×10−4)×6002=25×(5.00×10−4)×(3.60×105)=25×180=4500
• Denominator: $2 \times 0.500^2 = 2 \times 0.250 = 0.500$2×0.5002=2×0.250=0.500

Therefore, the coefficient $A$A is as follows: $A = \frac{4500}{0.500} = 9000$A=0.5004500​=9000

The natural number we are looking for is $9000$9000.

Answer: 9000