The acceleration of an electron at a moment in a magnetic field $\vec{B} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ is $\vec{a} = x\hat{i} - 2\hat{j} + \hat{k}$. The value of $x$ is

An electron is moving at a speed of $3.2 \times 10^7 \ m/s$ in a magnetic field of $12 \times 10^{-4} \ T$ perpendicular to the direction of motion of the electron. The radius of the path of the electron is . . . . . . $cm$. $(e = 1.6 \times 10^{-19} \ C$ and $m_e = 9 \times 10^{-31} \ kg)$

$A$ uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields,then the electron

$A$ charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will be

$A$ proton with a kinetic energy of $2.0 \, eV$ moves into a region of uniform magnetic field of magnitude $\frac{\pi}{2} \times 10^{-3} \, T$. The angle between the direction of magnetic field and velocity of proton is $60^{\circ}$. The pitch of the helical path taken by the proton is $.......... \, cm$. (Take,mass of proton $= 1.6 \times 10^{-27} \, kg$ and charge on proton $= 1.6 \times 10^{-19} \, C$)

The ratio of time periods of $\alpha$-particle and proton moving on circular path in a uniform magnetic field is . . . . . . .