$A$ positively charged $(+q)$ particle of mass $m$ has kinetic energy $K$ and enters vertically downward into a horizontal magnetic field of induction $\vec{B}$. The acceleration of the particle is:

$A$ particle of charge $q = -16 \times 10^{-18} \, C$ moving with velocity $v = 10 \, m/s$ along the $x$-axis enters a region where a magnetic field of induction $B$ is along the $y$-axis,and an electric field of magnitude $E = 10^4 \, V/m$ is along the negative $z$-axis. If the charged particle continues moving along the $x$-axis,the magnitude of $B$ is:

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

$A$ beam of protons moving with a velocity $1.6 \times 10^5 \ m/s$ enters a uniform magnetic field of $\frac{\pi}{10} \ T$ at an angle $60^{\circ}$ to the direction of the field. The pitch of the helical path of the protons is (mass of proton $= 1.6 \times 10^{-27} \ kg$)

$A$ proton of velocity $(3\hat i + 2\hat j) \, ms^{-1}$ enters a magnetic field of $(2\hat j + 3\hat k) \, T$. The acceleration produced in the proton is (charge to mass ratio of proton $= 0.96 \times 10^8 \, C/kg$)

An electron experiences a force $\vec{F} = (4.0\,\hat{i} + 3.0\,\hat{j}) \times 10^{-13}\,N$ in a uniform magnetic field when its velocity is $\vec{v}_1 = 2.5\,\hat{k} \times 10^7\,m/s$. When the velocity is redirected to $\vec{v}_2 = (1.5\,\hat{i} - 2.0\,\hat{j}) \times 10^7\,m/s$,the magnetic force on the electron is zero. The magnetic field $\vec{B}$ is: