Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and…
Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields $\vec{E}=E_{0} \hat{j}$ and $\vec{B}=B_{0} \hat{j}$. At time $t=0$, this charge has velocity $\vec{v}$ in the in the $x$-y plane, making an angle $\theta$ with the $x$-axis. Which of the following option(s) is (are) correct for time $t>0$ ?
If $\theta=0^{\circ}$, the charge moves in a circular path in the $x$ - $z$ plane.
If $\theta=0^{\circ}$, the charge undergoes helical motion with constant pitch along the $y$-axis.
If $\theta=10^{\circ}$, the charge undergoes helical motion with its pitch increasing with time, along the $y$-axis.
If $\theta=90^{\circ}$, the charge undergoes linear but accelerated motion along the $y$-axis.
Solution
If $\theta=0^{\circ}$, the charged particle is projected along $x$-axis, due to magnetic field, $\vec{B}$ the charged particle will tend to move in a circular path in $y-z$ plane but due to force of electric field $\vec{E}$, the particle will move in a helical path with increasing pitch. Hence options (A) and (B) are wrong. If $\theta=10^{\circ}$, we can resolve velocity into two rectangular components. One along $x$-axis $\left(v \cos 10^{\circ}\right)$ and one along $y$-axis $\left(v \sin 10^{\circ}\right)$. Due to $v \cos 10^{\circ}$, the particle will move in circular path and due to $v \sin 10^{\circ}$ plus the force due to electric field, the particle will undergo helical motion with its pitch increasing.
If $\theta=90^{\circ}$, the charge is moving along the magnetic field. Therefore the force due to magnetic field is zero. But the force due to electric field will accelerate the particle along $y$-axis.
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