$A$ strong magnetic field is applied on a stationary electron. Then,the electron

$A$ 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 $= 0.96 \times 10^8 \ C/kg$)

The radius of the path of an electron moving at a speed of $3.2 \times 10^7 \ m/s$ in a magnetic field of $6 \times 10^{-4} \ T$ perpendicular to it is (mass of electron is $9 \times 10^{-31} \ kg$ and charge of electron is $1.6 \times 10^{-19} \ C$). (in $cm$)

$A$ nucleus at rest undergoes $\alpha -$ decay according to the equation ${}_{92}^{226}X \to Y + \alpha$. At $t=0$,the emitted $\alpha -$ particle enters a region of space where a uniform magnetic field $\vec{B} = B_0 \hat{i}$ and an electric field $\vec{E} = E_0 \hat{i}$ exist. The $\alpha -$ particle enters the region with velocity $\vec{v} = v_0 \hat{j}$ at $x = 0$. After a time $t = \left( \sqrt{3} \times 10^7 \frac{m_{\alpha}}{q_{\alpha} E_0} \right)$,the particle is found to have twice its initial speed. The initial speed $v_0$ of the $\alpha -$ particle 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:

$A$ particle of charge $-q$ and mass $m$ enters a uniform magnetic field $\vec{B}$ at $A$ with speed $v_1$ at an angle $\alpha$ and leaves the field at $C$ with speed $v_2$ at an angle $\beta$ as shown. Then