A charged particle moves along circular path in a uniform magnetic field in a cyclotron. The kinetic energy of the charged particle increases to $4$ times its initial value. What will be the ratio of new radius to the original radius of circular path of the charged particle

A charged particle of charge $q$ and mass $m$, gets deflected through an angle $\theta$ upon passing through a square region of side $a$, which contains a uniform magnetic field $B$ normal to its plane. Assuming that the particle entered the square at right angles to one side, what is the speed of the particle?

Two parallel wires in the plane of the paper are distance $X _0$ apart. A point charge is moving with speed $u$ between the wires in the same plane at a distance $X_1$ from one of the wires. When the wires carry current of magnitude $I$ in the same direction, the radius of curvature of the path of the point charge is $R_1$. In contrast, if the currents $I$ in the two wires have direction opposite to each other, the radius of curvature of the path is $R_2$.

If $\frac{x_0}{x_1}=3$, the value of $\frac{R_1}{R_2}$ is.

An electron enters a chamber in which a uniform magnetic field is present as shown below.  An electric field of appropriate magnitude is also applied, so that the electron travels undeviated without any change in its speed through the chamber. We are ignoring gravity. Then, the direction of the electric field is

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 $q_p = 1.6\times10^{-19}\, C)$