A particle of mass $m$ and charge $q$ moves with a constant velocity $v$ along the positive $x$ direction. It enters a region containing a uniform magnetic field $B$ directed along the negative $z$ direction, extending from $x = a$ to $x = b$. The minimum value of $v$ required so that the particle can just enter the region $x > b$ is

An electron with kinetic energy $5 \mathrm{eV}$ enters a region of uniform magnetic field of $3 \mu \mathrm{T}$ perpendicular to its direction. An electric field $\mathrm{E}$ is applied perpendicular to the direction of velocity and magnetic field. The value of $\mathrm{E}$, so that electron moves along the same path, is . . . . . $\mathrm{NC}^{-1}$.

(Given, mass of electron $=9 \times 10^{-31} \mathrm{~kg}$, electric charge $=1.6 \times 10^{-19} \mathrm{C}$ )

A proton is projected with a velocity $10^7\, m/s$, at right angles to a uniform magnetic field of induction $100\, mT$. The time (in second) taken by the proton to traverse $90^o$ arc is  $(m_p = 1.65\times10^{-27}\, kg$ and $q_p = 1.6\times10^{-19}\, C)$

The figure shows a region of length $'l'$ with a uniform magnetic field of $0.3\, T$ in it and a proton entering the region with velocity $4 \times 10^{5}\, ms ^{-1}$ making an angle $60^{\circ}$ with the field. If the proton completes $10$ revolution by the time it cross the region shown, $l$ is close to....... $m$

(mass of proton $=1.67 \times 10^{-27} \,kg ,$ charge of the proton $\left.=1.6 \times 10^{-19}\, C \right)$

An electron enters a region where electrostatic field is $20\,N/C$ and magnetic field is $5\,T$. If electron passes undeflected through the region, then velocity of electron will be.....$m{s^{ - 1}}$

In the given figure, the electron enters into the magnetic field. It deflects in ...... direction