$A$ charged particle is moving in a uniform magnetic field $(2 \hat{i} + 3 \hat{j}) \text{ T}$. If it has an acceleration of $(\alpha \hat{i} - 4 \hat{j}) \text{ m/s}^2$,then the value of $\alpha$ will be:

An electron accelerated through a potential difference $V$ passes through a uniform transverse magnetic field and experiences a force $F$. If the accelerating potential is increased to $2V$,the electron in the same magnetic field will experience a force:

In the $xy$-plane,the region $y > 0$ has a uniform magnetic field $B_1 \hat{k}$ and the region $y < 0$ has another uniform magnetic field $B_2 \hat{k}$. $A$ positively charged particle is projected from the origin along the positive $y$-axis with speed $v_0 = \pi \text{ m s}^{-1}$ at $t = 0$,as shown in the figure. Neglect gravity in this problem. Let $t = T$ be the time when the particle crosses the $x$-axis from below for the first time. If $B_2 = 4 B_1$,the average speed of the particle,in $\text{m s}^{-1}$,along the $x$-axis in the time interval $T$ is. . . . . .

$A$ particle having charge of $1 \, C$,mass $1 \, kg$ and speed $1 \, m/s$ enters a uniform magnetic field,having magnetic induction of $1 \, T$,at an angle $\theta = 30^\circ$ between the velocity vector and the magnetic induction. The pitch of its helical path is (in meters):

An electron at rest is accelerated by a potential $V_{1}$ and then enters a uniform magnetic field,experiencing a force $F_{1}$. When the potential is changed to $V_{2}$,the force experienced by the electron becomes $2F_{1}$. The ratio of $V_{1}$ to $V_{2}$ is:

$A$ strong magnetic field is applied on a stationary electron,then