$A$ proton of velocity $(3\hat i + 2\hat j) \, ms^{-1}$ enters a magnetic field of $(2\hat j + 3\hat k) \, T$. The acceleration produced in the proton is (charge to mass ratio of proton $= 0.96 \times 10^8 \, C/kg$)

In the $xy$-plane,the region $y > 0$ has a uniform magnetic field $B_1 \hat{k}$ and the region $y < 0$ has another uniform magnetic field $B_2 \hat{k}$. $A$ positively charged particle is projected from the origin along the positive $y$-axis with speed $v_0 = \pi \text{ m s}^{-1}$ at $t = 0$,as shown in the figure. Neglect gravity in this problem. Let $t = T$ be the time when the particle crosses the $x$-axis from below for the first time. If $B_2 = 4 B_1$,the average speed of the particle,in $\text{m s}^{-1}$,along the $x$-axis in the time interval $T$ is. . . . . .

An electron enters the space between the plates of a charged capacitor as shown. The charge density on the plate is $\sigma$. Electric intensity in the space between the plates is $E$. $A$ uniform magnetic field $B$ also exists in that space perpendicular to the direction of $E$. The electron moves perpendicular to both $\vec{E}$ and $\vec{B}$ without any change in direction. The time taken by the electron to travel a distance $\ell$ in the space is

$A$ 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

$A$ particle of mass $m$ and charge $q$ enters a region of magnetic field (as shown) with speed $v$. There is a region in which the magnetic field is absent,as shown. The particle after entering the region collides elastically with a rigid wall. The time after which the velocity of the particle becomes anti-parallel to its initial velocity is

Electrons move at right angles to a magnetic field of $1.5 \times 10^{-2} \text{ T}$ with a speed of $6 \times 10^7 \text{ m/s}$. If the specific charge of the electron is $1.7 \times 10^{11} \text{ C/kg}$,the radius of the circular path will be...... $\text{cm}$.