Question

A particle of mass m and charge q is released from the origin in a region occupied by electric field E and magnetic field B, such that $\overrightarrow{B}=-B_0\hat{j};\overrightarrow{E}=E_0\hat{k}$.

If the subsequent equation of motion of the particle in z direction is given by $z=a\ast R[1-cos\omega t]$. Find a.

Answer 1

In fig shown $B_0$ in outward. As soon as we release the particle at origin. Electric field will apply force $F_e=q{E}_0$ on the particle. Due to this force, the charge particle will gain acceleration and hence velocity along z-axis. And as it gains velocity along z-axis, then magnetic field will apply the force $F_b$ on charge particle. Now both the forces are always in x-z plane, so path of charge will be in x-z plane as shown. Let at any time its velocity is $\overrightarrow{v}=v_x\hat{i}+v_z\hat{k}$. Let $\overrightarrow{a}=a_x\hat{i}+a_z\hat{k}$ is the acceleration at any time.
Then ${\overrightarrow{F}}_{net}=m\overrightarrow{a}$
$\Rightarrow$qE_0\hat k-q(v_x\hat i+v_z\hat k)\times B_0\hat j=m(a_x\hat i+a_z\hat k)\Rightarrow $q{E}_0\hat{k}-q{v}_x{B}_0\hat{k}+q{v}_z{B}_0\hat{i}=m(a_x\hat{i}+a_x\hat{k})$
$\Rightarrow m{a}_x=q{v}_z{B}_0$ and $m{a}_z=q(E_0-v_x{B}_0)$ (i)
$\Rightarrow m\frac{d{v}_x}{dt}=q{v}_z{B}_0$ and $m\frac{d{v}_z}{dt}=q(E_0-v_x{B}_0)$ (ii)
From Eq. (ii), $\frac{d{v}_z}{dt}=\frac{q}{m}(E_0-v_x{B}_0)\Rightarrow \frac{d^2{v}_z}{d{t}^2}=\frac{-q{B}_0}{m}\frac{d{v}_x}{dt}$
$\Rightarrow \frac{d^2{v}_z}{d{t}^2}=\frac{-q{B}_0}{m}\frac{q{B}_0}{m}{v}_z=-{\left(\frac{q{B}_0}{m}\right)}^2{v}_z$
Solution of the above equation is: $v_z=v_{z0}sin(\omega t+\varphi )$ (iii)
$(where\omega =\frac{q{B}_0}{m})$
Putting at $t=0,v_z=0$, we get $\varphi =0$
Differentiating Eq. (iii),
$\frac{d{v}_z}{dt}=v_{z0}\omega cos(\omega t+\varphi )\Rightarrow a_z=\omega v_{z0}cos\omega t$
at $t=0,a_z=\frac{F_e}{m}=\frac{q{E}_0}{m}$
$\Rightarrow \frac{q{E}_0}{m}=\frac{q{B}_0}{m}{v}_{z0}cos{0}^o\Rightarrow v_{z0}=\frac{E_0}{B_0}$
So Eq. (iii) can be written as: $v_z=\frac{E_0}{B_0}sin\left(\omega t\right)$ (iv)
Putting the value of $v_z$ in Eq. (ii), $m\frac{d{v}_x}{dt}=\frac{q{E}_0}{B_0}\left(sin\omega t\right)B_0$
$\Rightarrow {\int }_0^{v_x}d{v}_x={\int }_0^t\frac{q{E}_0}{m}sin\omega tdt\Rightarrow v_x=\frac{q{E}_0}{m\omega }[1-cos\omega t]$
$\Rightarrow v_x=\frac{qE}{m\left(q{B}_0/m\right)}[1-cos\omega t]\Rightarrow v_x=\frac{E_0}{B_0}[1-cos\omega t]$ (v)
Equations (iv) and (v) give velocity as a function of time.
From Eq. (v), $\frac{dx}{dt}=\frac{E_0}{B_0}(1-cos\omega t)$
$\Rightarrow {\int }_0^{x}dx=\frac{E_0}{B_0}{\int }_0^t(1-cos\omega t)dt$
$\Rightarrow x=\frac{E_0}{\omega B_0}[\omega t-sin\omega t]$ (vi)
From Eq. (iv)
$\frac{dz}{dt}=\frac{E_0}{B_0}sin\omega t\Rightarrow {\int }_0^{z}dz=\frac{E_0}{B_0}{\int }_0^{t}sin\omega tdt$
$\Rightarrow z=\frac{E_0}{\omega B_0}[1-cos\omega t]$ (vii)
Equations (vi) and (vii) give the path of the verticle.
We see that at $t=2\pi$
$x=\frac{E_0}{B_0}2\pi ,z=0$
This path is known as cycloidal path. This is the same kind of path as followed by a point on the rim of a purely rolling wheel, which is explained as below.
Let a wheel is rolling purely with angular velocity $\omega$ as shown. Let at $t=0$, a point P on the rim is at the bottom-most position of the wheel. After time t, the wheel turns by an angle $\theta$ as shown and point P comes at point P'. Let us find the coordinates of P'.
$x=R\theta -Rsin\theta =R\omega t-Rsin\omega t=R[\omega t-sin\omega t]$ (viii)
$z=R-Rcos\theta =R[1-cos\omega t]$ (ix)
Equation (viii) and (ix) resemble Eqs (vi) and (vii).
Hence, the resulting path followed by charge is cycloidal.