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 speed of the particle as a function of the z-coordinate is $v=\sqrt{\frac{xq{E}_{0}z}{m}}$. Find $x.$

Answer 1

To find velocity as function of Z-coordinates.
From equations (iv) and (v):
$v=\sqrt{v_x^2+v_z^2}=\frac{E_0}{B_0}\sqrt{(1-cos\omega t{)}^2+(sin\omega t{)}^2}$
$=\frac{E_0}{B_0}\sqrt{(1-cos\omega t{)}^2+(1-cos\omega t\left)\right(1+cos\omega t)}$
$=\frac{E_0}{B_0}\sqrt{(1-cos\omega t{)}^2}=\frac{E_0}{B_0}\sqrt{\frac{z\omega B_0}{E_0}2}$ ..... [From Eq. (vii)]
$=\sqrt{\frac{E_0^2}{B_0^2}\frac{zq{B}_0}{m}\frac{B_0}{E_0}2}=\sqrt{\frac{2q{E}_{0}z}{m}}$
Alternate method to find velocity: Since the magnetic field does not perform any work, therefore, whatever has been the gain in kinetic energy it is only because of the work done by electric field. Applying work-energy theorem, $W_E=\Delta K$
$\Rightarrow q{E}_{0}z=\frac{1}{2}m{v}^2-0$

$v=\sqrt{\frac{2q{E}_{0}z}{m}}$