Work done by a magnetic field of $1\,T$ on a particle of charge $1\,C$ moving with a speed of $5\,m/s$ in one second is........$J$.

$A$ proton of velocity $v = (3 \hat{i} + 2 \hat{j}) \ m/s$ enters a magnetic field of induction $B = (2 \hat{j} + 3 \hat{k}) \ T$. The acceleration produced in the proton in $m/s^2$ is (Specific charge of proton $= 0.96 \times 10^8 \ C/kg$)

$A$ charged particle moves with some initial velocity along the direction of an external magnetic field $B$. Now, if we apply a uniform electric field $E$ perpendicular to the magnetic field, then the trajectory of the charged particle will be

An electron accelerated through a potential difference $V$ passes through a uniform transverse magnetic field and experiences a force $F$. If the accelerating potential is increased to $2V$,the electron in the same magnetic field will experience a force:

An electron,a proton,a deuteron,and an alpha particle,each having the same speed,are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radii of the circular orbits of these particles are respectively $R_e, R_p, R_d$,and $R_\alpha$. It follows that:

Given below are two statements:
Statement $I$: The electric force changes the speed of the charged particle and hence changes its kinetic energy; whereas the magnetic force does not change the kinetic energy of the charged particle.
Statement $II$: The electric force accelerates the positively charged particle perpendicular to the direction of the electric field. The magnetic force accelerates the moving charged particle along the direction of the magnetic field.
In the light of the above statements,choose the most appropriate answer from the options given below: