A particle of mass $m = 1.67 \times 10^{-27}\, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $1$ $tesla$ along the direction shown in the figure. the particle leaves the magnetic field at point $D,$ then the distance $CD$ is :-

An electron enters a region where electrostatic field is $20\,N/C$ and magnetic field is $5\,T$. If electron passes undeflected through the region, then velocity of electron will be.....$m{s^{ - 1}}$

An electron moving with a velocity ${\vec V_1} = 2\,\hat i\,\, m/s$ at a point in a magnetic field experiences a force ${\vec F_1} =  - 2\hat j\,N$ .  If the electron is moving with a velocity ${\vec V_2} = 2\,\hat j \,\,m/s$ at the same point, it experiences a force ${\vec F_2} =  + 2\,\hat i\,N$ .  The force the electron would experience if it were moving with a velocity ${\vec V_3} = 2\hat k$  $m/s$ at the same point is

Which law is useful to determine relation between current and magnetic fields due to it.

A proton of velocity $\left( {3\hat i + 2\hat j} \right)\,ms^{-1}$ enters a magnetic field of  $(2\hat j + 3\hat k)\, tesla$. The acceleration produced in the proton is (charge to mass ratio of proton $= 0.96 \times10^8\,Ckg^{-1}$)

A proton (mass $ = 1.67 \times {10^{ - 27}}\,kg$ and charge $ = 1.6 \times {10^{ - 19}}\,C)$ enters perpendicular to a magnetic field of intensity $2$ $weber/{m^2}$ with a velocity $3.4 \times {10^7}\,m/\sec $. The acceleration of the proton should be