$A$ particle having charge of $1 \, C$,mass $1 \, kg$ and speed $1 \, m/s$ enters a uniform magnetic field,having magnetic induction of $1 \, T$,at an angle $\theta = 30^\circ$ between the velocity vector and the magnetic induction. The pitch of its helical path is (in meters):

In a chamber,a uniform magnetic field of $6.5 \; G$ $(1 \; G = 10^{-4} \; T)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \; m s^{-1}$ normal to the field. The radius of the circular orbit of the electron is $4.2 \; cm$. Obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain. $(e = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$

$A$ particle of specific charge (charge/mass) $\alpha$ starts moving from the origin under the action of an electric field $\vec{E} = E_0 \hat{i}$ and a magnetic field $\vec{B} = B_0 \hat{k}$. Its velocity at $(x_0, y_0, 0)$ is $(4 \hat{i} + 3 \hat{j})$. The value of $x_0$ is:

An electron is travelling horizontally towards the east. $A$ magnetic field in the vertically downward direction exerts a force on the electron along:

At $t = 0$,a charge $q$ is at the origin and moving in the $y$-direction with velocity $\vec{v} = v\hat{j}$. The charge moves in a magnetic field that is for $y > 0$ out of the page and given by $B_1\hat{k}$ and for $y < 0$ into the page and given by $-B_2\hat{k}$. The charge's subsequent trajectory is shown in the sketch. From this information,we can deduce that:

$A$ rod $AB$ moves with a uniform velocity $v$ in a uniform magnetic field as shown in the figure.