$A$ particle of charge $q$ and mass $m$ starts moving from the origin under the action of an electric field $\vec E = E\hat i$ and a magnetic field $\vec B = B\hat i$ with an initial velocity $\vec v = v_0\hat j$. The speed of the particle will become $2v_0$ after a time:

$A$ charged particle is moving along a magnetic field line. What is the magnetic force acting on the particle? $(\sin 0^{\circ}=0, \sin \frac{\pi}{2}=1)$

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. $(e = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$

$A$ proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of $2 \times 10^5 \text{ ms}^{-1}$. When the electric field is switched off,the proton moves along a circular path of radius $2 \text{ cm}$. The magnitude of the electric field is $x \times 10^4 \text{ N/C}$. The value of $x$ is . . . . . . . (Take the mass of the proton $= 1.6 \times 10^{-27} \text{ kg}$ and charge $e = 1.6 \times 10^{-19} \text{ C}$)

In a region of space,a uniform magnetic field $B$ exists in the $y-$direction. $A$ proton is fired from the origin,with its initial velocity $v$ making a small angle $\alpha$ with the $y-$direction in the $yz$ plane. In the subsequent motion of the proton,

$A$ charged particle moving along a straight line path enters a uniform magnetic field of $4 \ mT$ at right angles to the direction of the magnetic field. If the specific charge of the charged particle is $8 \times 10^7 \ C \ kg^{-1}$,the angular velocity of the particle in the magnetic field is