$A$ charge of $4.0 \mu C$ is moving with a velocity of $4.0 \times 10^6 \ m/s$ along the positive $y$-axis under a magnetic field $\vec{B}$ of strength $(2 \hat{k}) \ T$. The force acting on the charge is $x \hat{i} \ N$. The value of $x$ is . . . . . . .

$A$ 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:

$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ along the direction shown in the figure. The particle leaves the magnetic field at point $D$. Calculate the distance $CD$. (Assume the velocity of the particle $v = 10^7 \, m/s$)

Electrons move at right angles to a magnetic field of $1.5 \times 10^{-2} \text{ T}$ with a speed of $6 \times 10^7 \text{ m/s}$. If the specific charge of the electron is $1.7 \times 10^{11} \text{ C/kg}$,the radius of the circular path will be...... $\text{cm}$.

$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ as shown in the figure. The magnetic field is directed into the plane of the paper. The particle enters at point $C$ with an angle of $45^{\circ}$ with the boundary. Find the time spent by the particle in the magnetic field region in $ns$.

Two charged particles $A$ and $B$ of masses $m$ and $2m$,charges $2q$ and $3q$ respectively,moving with the same velocity,enter a uniform magnetic field such that both particles make the same angle $( < 90^{\circ} )$ with the direction of the magnetic field. Then the ratio of the pitches of the helical paths of the particles $A$ and $B$ is