$A$ proton moving with a velocity $2.5 \times 10^7 \ m/s$ enters a magnetic field of intensity $2.5 \ T$ making an angle $30^{\circ}$ with the magnetic field. The force on the proton is

Which of the following statements is true?

An electron having a charge $e$ moves with a velocity $\vec{v}$ in the positive $x$-direction. $A$ magnetic field $\vec{B}$ acts on it in the positive $y$-direction. The force on the electron acts in (where the outward direction is taken as the positive $z$-axis):

An electron enters a chamber in which a uniform magnetic field is present as shown below. An electric field of appropriate magnitude is also applied,so that the electron travels undeviated without any change in its speed through the chamber. We are ignoring gravity. Then,the direction of the electric field is

$A$ proton is moving perpendicular to a uniform magnetic field of $2.5 \ T$ with $2 \ MeV$ kinetic energy. The force on the proton is . . . . . . $N$. (Mass of proton $= 1.6 \times 10^{-27} \ kg$,charge of proton $= 1.6 \times 10^{-19} \ C$)

An electron is accelerated by a potential difference of $12000\, V$. It then enters a uniform magnetic field of $10^{-3}\, T$ applied perpendicular to the path of the electron. Find the radius of the path. Given: mass of electron $m = 9 \times 10^{-31}\, kg$ and charge on electron $q = 1.6 \times 10^{-19}\, C$.