$A$ particle of charge $1.0 \times 10^{-16} \ C$ moves through a uniform magnetic field $B=B_0(\hat{i}+4 \hat{j}) \ T$. The particle velocity at some instant is $v=(2 \hat{i}+4 \hat{j}) \ ms^{-1}$ and the magnetic force acting on it is $3 \times 10^{-16} \hat{k} \ N$. The magnitude of $B_0$ is (in $T$)

Under the influence of a uniform magnetic field,a charged particle is moving in a circle of radius $R$ with constant speed $v$. The time period of the motion

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

An $\alpha$-particle of $1 \, MeV$ energy moves on a circular path in a uniform magnetic field. Then,the kinetic energy of a proton in the same magnetic field for a circular path of double radius is ...... $MeV$.

Two long parallel conductors $S_{1}$ and $S_{2}$ are separated by a distance $10 \, cm$ and carry currents of $4 \, A$ and $2 \, A$ respectively. The conductors are placed along the $x$-axis in the $X-Y$ plane. There is a point $P$ located between the conductors (as shown in the figure). $A$ charged particle of $3 \pi \, C$ is passing through the point $P$ with velocity $\overrightarrow{v} = (2 \hat{i} + 3 \hat{j}) \, m/s$; where $\hat{i}$ and $\hat{j}$ represent unit vectors along the $x$ and $y$ axes respectively. The force acting on the charged particle is $4 \pi \times 10^{-5} (-x \hat{i} + 2 \hat{j}) \, N$. The value of $x$ is:

Two electrons,$e_1$ and $e_2$ of mass $m$ and charge $q$ are injected into the perpendicular direction of the magnetic field $B$ such that the kinetic energy of $e_1$ is double than that of $e_2$. The relation of their frequencies of rotation,$f_1$ and $f_2$ is