The acceleration of an electron at a moment in a magentic 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

A proton of mass $1.67 \times {10^{ - 27}}\,kg$ and charge $1.6 \times {10^{ - 19}}\,C$ is projected with a speed of $2 \times {10^6}\,m/s$ at an angle of $60^\circ $ to the $X - $ axis. If a uniform magnetic field of $0.104$ $Tesla$ is applied along $Y - $ axis, the path of proton is

An electron having charge $1.6 \times {10^{ - 19}}\,C$ and mass $9 \times {10^{ - 31}}\,kg$ is moving with $4 \times {10^6}\,m{s^{ - 1}}$ speed in a magnetic field $2 \times {10^{ - 1}}\,tesla$ in a circular orbit. The force acting on electron and the radius of the circular orbit will be

A magnetic field $\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{\mathrm{j}}$ exists in the region $\mathrm{a} < \mathrm{x} < 2 \mathrm{a}$ and $\vec{B}=-B_0 \hat{j}$, in the region $2 \mathrm{a} < \mathrm{x} < 3 \mathrm{a}$, where $\mathrm{B}_0$ is a positive constant. $\mathrm{A}$ positive point charge moving with a velocity $\overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{\dot{i}}$, where $v_0$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like,

An electron is moving along the positive $X$$-$axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative $X$$-$axis. This can be done by applying the magnetic field along

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 $=1.6 \times 10^{-19}\,kg)$