$A$ particle of charge $q$ moves with a velocity $\vec{V} = a \hat{i}$ in a magnetic field $\vec{B} = b \hat{j} + c \hat{k}$,where $a$,$b$,and $c$ are constants. The magnitude of the force experienced by the particle is:

Two charged particles,having the same kinetic energy,are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of the radii of their circular paths is $6:5$ and their respective masses ratio is $9:4$,then the ratio of their charges will be.

$A$ 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?

Under the influence of a uniform magnetic field,a charged particle is moving in a circle of radius $R$ with constant speed $v$. The time period of the motion

$A$ proton moving in a perpendicular magnetic field possesses energy $E$. The strength of the magnetic field is increased four times, but the proton is constrained to move in a path of the same radius. The kinetic energy of the proton will increase by: (in $\times$)

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: $1 \; eV = 1.6 \times 10^{-19} \; J$)