$A$ horizontal conductor is oriented north-south and carries some current. $A$ positively charged particle located vertically above it and having a velocity directed northward experiences an upward force. What is the direction of the force if this charged particle were located to the east of the conductor and had a velocity directed towards the conductor?

$A$ charged particle is moving with velocity $v$ in a magnetic field of induction $B$. The force on the particle will be maximum when

$A$ positive charge particle having charge $q$ and mass $m$ has velocity $\vec v = v\left( {\frac{{\hat i + \hat k}}{{\sqrt 2 }}} \right)$ in the magnetic field $\vec B = B\hat j$ and electric field $\vec E = E\hat j$ at the origin. Its speed as a function of $y$ is:

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. . . . . .

The region between $y = 0$ and $y = d$ contains a magnetic field $\vec B = B\hat k$. $A$ particle of mass $m$ and charge $q$ enters the region with a velocity $\vec v = v\hat i$. If $d = \frac{mv}{2qB}$,the acceleration of the charged particle at the point of its emergence at the other side is

Two charged particles of mass $m$ and charge $q$ each are projected from the origin simultaneously with the same speed $V$ in a transverse magnetic field $B$. If $\vec{r}_1$ and $\vec{r}_2$ are the position vectors of the particles (with respect to the origin) at $t = \frac{\pi m}{qB}$, then the value of $\vec{r}_1 \cdot \vec{r}_2$ at that time is: