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

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. . . . . .

$A$ charged particle experiences a magnetic force in the presence of a magnetic field. Which of the following statements is correct?

$A$ particle of mass $0.6 \,g$ and having charge of $25 \,nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4 \,ms^{-1}$ in a uniform magnetic field. The value of the magnetic induction is $\left(g=10 \,ms^{-2}\right)$.

$A$ proton of energy $200\, MeV$ enters a magnetic field of $5\, T$. If the direction of the field is from south to north and the motion is upward,the force acting on it will be:

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