A current $i$ ampere flows along the inner conductor of a coaxial cable and returns along the outer conductor of the cable, then the magnetic induction at any point outside the conductor at a distance $r$ metre from the axis is
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$(A)$ decreasing the number of turns
$(B)$ increasing the magnetic field
$(C)$ decreasing the area of the coil
$(D)$ decreasing the torsional constant of the spring
Choose the most appropriate answer from the options given below.
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column $I$
column $II$ column $III$ $(I)$ Electron with $\overrightarrow{\mathrm{v}}=2 \frac{\mathrm{E}_0}{\mathrm{~B}_0} \hat{\mathrm{x}}$ $(i)$ $\overrightarrow{\mathrm{E}}=\mathrm{E}_0^2 \hat{\mathrm{Z}}$ $(P)$ $\overrightarrow{\mathrm{B}}=-\mathrm{B}_0 \hat{\mathrm{x}}$ $(II)$ Electron with $\overrightarrow{\mathrm{v}}=\frac{\mathrm{E}_0}{\mathrm{~B}_0} \hat{\mathrm{y}}$ $(ii)$ $\overrightarrow{\mathrm{E}}=-\mathrm{E}_0 \hat{\mathrm{y}}$ $(Q)$ $\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{\mathrm{x}}$ $(III)$ Proton with $\overrightarrow{\mathrm{v}}=0$ $(iii)$ $\overrightarrow{\mathrm{E}}=-\mathrm{E}_0 \hat{\mathrm{x}}$ $(R)$ $\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{\mathrm{y}}$ $(IV)$ Proton with $\overrightarrow{\mathrm{v}}=2 \frac{\mathrm{E}_0}{\mathrm{~B}_0} \hat{\mathrm{x}}$ $(iv)$ $\overrightarrow{\mathrm{E}}=\mathrm{E}_0 \hat{\mathrm{x}}$ $(S)$ $\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{\mathrm{z}}$ ($1$) In which case will the particle move in a straight line with constant velocity?
$[A] (II) (iii) (S)$ $[B] (IV) (i) (S)$ $[C] (III) (ii) (R)$ $[D] (III) (iii) (P)$
($2$) In which case will the particle describe a helical path with axis along the positive $z$ direction?
$[A] (II) (ii) (R)$ $[B] (IV) (ii) (R)$ $[C] (IV) (i) (S)$ $[D] (III) (iii)(P)$
($3$) In which case would be particle move in a straight line along the negative direction of y-axis (i.e., more along $-\hat{y}$ )?
$[A] (IV) (ii) (S)$ $[B] (III) (ii) (P)$ $[C]$ (II) (iii) $(Q)$ $[D] (III) (ii) (R)$
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