$A$ proton moving with a velocity $2.5 \times 10^7 \ m/s$ enters a magnetic field of intensity $2.5 \ T$ making an angle $30^{\circ}$ with the magnetic field. The force on the proton is

$A$ 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:

When a charged particle moving with velocity $\vec{V}$ is subjected to a magnetic field of induction $\vec{B}$,the force on it is non-zero. This implies that the:

$A$ beam of well-collimated cathode rays travelling with a speed of $5 \times 10^6 \, m/s$ enters a region of mutually perpendicular electric and magnetic fields and emerges undeviated from this region. If $|B| = 0.02 \, T$,the magnitude of the electric field is:

Two particles $X$ and $Y$ having equal charges,are accelerated through the same potential difference. They enter a region of a uniform magnetic field and describe a circular path of radii $r_1$ and $r_2$ respectively. The ratio of the mass of $X$ to that of $Y$ is

An electron gun with its collector at a potential of $100 \; V$ fires out electrons in a spherical bulb containing hydrogen gas at low pressure $(\sim 10^{-2} \; mm$ of $Hg)$. $A$ magnetic field of $2.83 \times 10^{-4} \; T$ curves the path of the electrons in a circular orbit of radius $12.0 \; cm$. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons,and emitting light by electron capture; this method is known as the 'fine beam tube' method.) Determine $e/m$ from the data.