$A$ proton of mass $m$ and charge $+e$ is moving in a circular orbit in a magnetic field with energy $1\, MeV$. What should be the energy of $\alpha$-particle (mass = $4m$ and charge = $+2e$) so that it can revolve in the path of the same radius?

$A$ charged particle moves through a magnetic field in a direction perpendicular to it. Then the

An infinitely long straight conductor carries a current of $5 \, A$ as shown. An electron is moving with a speed of $10^{5} \, m/s$ parallel to the conductor. The perpendicular distance between the electron and the conductor is $20 \, cm$ at an instant. Calculate the magnitude of the force experienced by the electron at that instant in $\times 10^{-20} \, N$.

Uniform magnetic fields of different strengths ($B_1$ and $B_2$),both normal to the plane of the paper,exist as shown in the figure. $A$ charged particle of mass $m$ and charge $q$,at the interface at an instant,moves into the region $2$ with velocity $v$ and returns to the interface. It continues to move into region $1$ and finally reaches the interface. What is the displacement of the particle during this movement along the interface? (Consider the velocity of the particle to be normal to the magnetic field and $B_2 > B_1$)

$A$ charged particle of charge $q$ and mass $m$ is accelerated through a potential difference of $V$ volts. It enters a region of orthogonal magnetic field $B$. The radius of its circular path will be:

If a proton enters perpendicularly into a magnetic field with velocity $v$,the time period of revolution is $T$. If the proton enters with velocity $2v$,what will be the time period?