A proton is projected with velocity $\overrightarrow{ V }=2 \hat{ i }$ in a region where magnetic field $\overrightarrow{ B }=(\hat{i}+3 \hat{j}+4 \hat{k})\; \mu T$ and electric field $\overrightarrow{ E }=10 \hat{ i } \;\mu V / m .$ Then find out the net acceleration of proton (in $m / s ^{2}$)

One proton beam enters a magnetic field of ${10^{ - 4}}$ $T$ normally, Specific charge = ${10^{11}}\,C/kg.$ velocity = ${10^7}\,m/s$. What is the radius of the circle described by it....$m$

A proton with a kinetic energy of $2.0\,eV$ moves into a region of uniform magnetic field of magnitude $\frac{\pi}{2} \times 10^{-3}\,T$. The angle between the direction of magnetic field and velocity of proton is $60^{\circ}$. The pitch of the helical path taken by the proton is $..........cm$ (Take, mass of proton $=1.6 \times 10^{-27}\,kg$ and Charge on proton $=1.6 \times 10^{-19}\,kg)$

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron

A particle of mass $m$ and charge $q$ enters a magnetic field $B$ perpendicularly with a velocity $v$, The radius of the circular path described by it will be

Two protons move parallel to each other, keeping distance $r$ between them, both moving with same  velocity $\vec V\,$. Then the ratio of the electric and magnetic force of interaction between them is