$A$ particle of mass $M$ and charge $q$ moves with a constant velocity $V$ in the positive $x$-direction. $A$ constant magnetic field $B$ exists in the negative $z$-direction between $x = a$ and $x = b$. Find the minimum value of $V$ such that the particle enters the region $x > b$.

$A$ $2 \, MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5 \, T$. The force on the proton is

$A$ 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?

An ion beam of specific charge $5 \times 10^7 \ C/kg$ enters a uniform magnetic field of $4 \times 10^{-2} \ T$ with a velocity $2 \times 10^5 \ m/s$ perpendicularly. The radius of the circular path of the ions in meters will be:

$A$ charge $q_0$ moving with velocity $\overrightarrow{v}$ in a magnetic field of induction $\overrightarrow{B}$ experiences a force $\overrightarrow{F}$. The angle between $\overrightarrow{v}$ and $\overrightarrow{B}$ is $\theta$. The speed of $q_0$ after one second will be

$A$ homogeneous electric field $\vec{E}$ and a uniform magnetic field $\vec{B}$ are pointing in the same direction. $A$ proton is projected with its velocity parallel to $\vec{E}$. It will