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

Assertion : $A$ charge,whether stationary or in motion,produces a magnetic field around it.
Reason : Moving charges produce only electric field in the surrounding space.

Match List-$I$ with List-$II$ and choose the correct answer from the options given below:
| List-$I$ ($Y$ vs $X$) | List-$II$ (Shape of Graph) |
| :--- | :--- |
| $(A)$ $Y$ = magnetic susceptibility, $X$ = magnetising field | $(I)$ Linear graph passing through origin |
| $(B)$ $Y$ = magnetic field, $X$ = distance from centre of a current carrying wire for $x < a$ (where $a$ = radius of wire) | $(II)$ Graph with a curve decreasing towards the axis |
| $(C)$ $Y$ = magnetic field, $X$ = distance from centre of a current carrying wire for $x > a$ (where $a$ = radius of wire) | $(III)$ Horizontal straight line graph |
| $(D)$ $Y$ = magnetic field inside solenoid, $X$ = distance from center | $(IV)$ Linear graph starting from origin |

Two protons move parallel to each other,keeping a distance $r$ between them,both moving with the same velocity $\vec{v}$. Then the ratio of the electric force to the magnetic force of interaction between them is:

Two electrons are moving along parallel lines in the same direction with the same velocity. They will:

$A$ long,straight wire is turned into a loop of radius $10\,cm$ (see figure). If a current of $8\,A$ is passed through the loop,then the value of the magnetic field and its direction at the centre $C$ of the loop shall be close to