$A$ very long straight wire carries a current $I$. At the instant when a charge $+Q$ at point $P$ has velocity $\vec{V}$,as shown,the force on the charge is

An electron revolves around the nucleus in a circular path with angular momentum $\vec{L}$. $A$ uniform magnetic field $\vec{B}$ is applied perpendicular to the plane of its orbit. If the electron experiences a torque $\vec{\tau}$, then

$A$ particle moving with velocity $v$ having specific charge $(q/m)$ enters a region of magnetic field $B$ having width $d = \frac{3mv}{5qB}$ at an angle of $53^{\circ}$ to the boundary of the magnetic field. Find the angle $\theta$ shown in the diagram.

$A$ charge having $q/m$ equal to $10^8 \, C/kg$ and with velocity $3 \times 10^5 \, m/s$ enters into a uniform magnetic field $0.3 \, T$ at an angle $30^{\circ}$ with the direction of the field. The radius of curvature will be ...... $cm$.

$A$ particle of charge $1.6 \ \mu C$ and mass $16 \ \mu g$ is present in a strong magnetic field of $6.28 \ T$. The particle is then fired perpendicular to the magnetic field. The time required for the particle to return to its original location for the first time is . . . . . . $s$. $(\pi = 3.14)$

What is the radius of the path of an electron (mass $m = 9 \times 10^{-31} \; kg$ and charge $q = 1.6 \times 10^{-19} \; C$) moving at a speed of $v = 3 \times 10^{7} \; m/s$ in a magnetic field of $B = 6 \times 10^{-4} \; T$ perpendicular to it? What is its frequency? Calculate its energy in $keV$. (Given: $1 \; eV = 1.6 \times 10^{-19} \; J$)