$A$ charged particle of $2\,\mu\,C$ accelerated by a potential difference of $100\,V$ enters a region of uniform magnetic field of magnitude $4\,mT$ at a right angle to the direction of the field. The charged particle completes a semicircle of radius $3\,cm$ inside the magnetic field. The mass of the charged particle is $........\times 10^{-18}\,kg$.

In a chamber,a uniform magnetic field of $6.5 \; G$ $(1 \; G = 10^{-4} \; T)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \; m s^{-1}$ normal to the field. The radius of the circular orbit of the electron is $4.2 \; cm$. Obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain. $(e = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$

Two infinitely long straight wires $A$ and $B$,each carrying current $I$,are placed on the $x$ and $y$-axes,respectively. The current in wires $A$ and $B$ flows along $-\hat{i}$ and $\hat{j}$ directions,respectively. The force on a charged particle having charge $q$,moving from position $r = d(\hat{i} + \hat{j})$ with velocity $v = v\hat{i}$ is:

Two ions having same mass have charges in the ratio $1: 2$. They are projected normally in a uniform magnetic field with their speeds in the ratio $2: 3$. The ratio of the radii of their circular trajectories is -

When a charged particle moving with velocity $\vec{v}$ is subjected to a magnetic field of induction $\vec{B}$,the force on it is non-zero. This implies that

$A$ proton is moving with a uniform velocity of $10^{6} \ m/s$ along the $Y$-axis,under the joint action of a magnetic field along the $Z$-axis and an electric field of magnitude $2 \times 10^{4} \ V/m$ along the negative $X$-axis. If the electric field is switched off,the proton starts moving in a circle. The radius of the circle is nearly: (Given: $\frac{e}{m}$ ratio for proton $= 10^{8} \ C/kg$) (in $m$)