Two very long straight parallel wires,parallel to the $y-$axis,carry currents $4I$ and $I$ along the $+y$ direction and $-y$ direction,respectively. The wires pass through the $x-$axis at the points $(d, 0, 0)$ and $(-d, 0, 0)$ respectively. The graph of the magnetic field $z-$component as one moves along the $x-$axis from $x=-d$ to $x=+d$ is best given by:

In order to make the parabolas formed by singly ionized ions in one spectrograph and doubly ionized ions in another Thomson's mass spectrograph coincide,the electric fields and magnetic fields are kept in the ratios $1 : 2$ and $3 : 2$ respectively. Then the ratio of the masses of the ions is

$A$ charged particle (electron or proton) is introduced at the origin $(x=0, y=0, z=0)$ with a given initial velocity $\overrightarrow{v}$. $A$ uniform electric field $\overrightarrow{E}$ and magnetic field $\vec{B}$ are given in columns $I, II$ and $III$, respectively. The quantities $E_0, B_0$ are positive in magnitude.

Column $I$	Column $II$	Column $III$
$(I)$ Electron with $\overrightarrow{v}=2 \frac{E_0}{B_0} \hat{x}$	$(i)$ $\overrightarrow{E}=E_0 \hat{z}$	$(P)$ $\overrightarrow{B}=-B_0 \hat{x}$
$(II)$ Electron with $\overrightarrow{v}=\frac{E_0}{B_0} \hat{y}$	$(ii)$ $\overrightarrow{E}=-E_0 \hat{y}$	$(Q)$ $\overrightarrow{B}=B_0 \hat{x}$
$(III)$ Proton with $\overrightarrow{v}=0$	$(iii)$ $\overrightarrow{E}=-E_0 \hat{x}$	$(R)$ $\overrightarrow{B}=B_0 \hat{y}$
$(IV)$ Proton with $\overrightarrow{v}=2 \frac{E_0}{B_0} \hat{x}$	$(iv)$ $\overrightarrow{E}=E_0 \hat{x}$	$(S)$ $\overrightarrow{B}=B_0 \hat{z}$

$(1)$ In which case will the particle move in a straight line with constant velocity?
$(2)$ In which case will the particle describe a helical path with axis along the positive $z$ direction?
$(3)$ In which case would the particle move in a straight line along the negative direction of $y$-axis (i.e., move along $-\hat{y}$)?

Choose the correct statement.

Two particles carrying equal charges move parallel to each other with the speed $150 \ km/s$. If $F_1$ and $F_2$ are magnetic and electric forces between two charged particles, then $\frac{|F_1|}{|F_2|}$ is (Let $\mu_0 \varepsilon_0 = \frac{1}{9 \times 10^{16}} \ s^2/m^2$)

Two parallel beams of electrons moving in the same direction produce a mutual force: