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.

Two charged particles $A$ and $B$,each of charge $+e$ and masses $12 \, amu$ and $13 \, amu$ respectively,follow a circular trajectory in chamber $X$ after passing through a velocity selector as shown in the figure. Both particles enter the velocity selector with a speed of $1.5 \times 10^6 \, ms^{-1}$. $A$ uniform magnetic field of strength $1.0 \, T$ is maintained within the chamber $X$ and in the velocity selector,directed into the plane ($-z$ direction).

The dimensional formula of $\frac{1}{2} \mu_0 H^2$ (where $\mu_0$ is the permeability of free space and $H$ is the magnetic field intensity) is:

$A$ compass needle free to turn in a horizontal plane is placed at the centre of a circular coil of $30$ turns and radius $12 \;cm$. The coil is in a vertical plane making an angle of $45^{\circ}$ with the magnetic meridian. When the current in the coil is $0.35 \;A$,the needle points west to east.
$(a)$ Determine the horizontal component of the earth's magnetic field at the location.
$(b)$ The current in the coil is reversed,and the coil is rotated about its vertical axis by an angle of $90^{\circ}$ in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the place to be zero.

In a thin rectangular metallic strip, a constant current $I$ flows along the positive $x$-direction, as shown in the figure. The length, width, and thickness of the strip are $\ell$, $w$, and $d$, respectively. A uniform magnetic field $\vec{B}$ is applied on the strip along the positive $y$-direction. Due to this, the charge carriers experience a net deflection along the $z$-direction. This results in the accumulation of charge carriers on the surface $PQRS$ and the appearance of equal and opposite charges on the face opposite to $PQRS$. A potential difference along the $z$-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross-section of the strip and carried by electrons.
$1.$ Consider two different metallic strips ($1$ and $2$) of the same material. Their lengths are the same, widths are $w_1$ and $w_2$, and thicknesses are $d_1$ and $d_2$, respectively. Two points $K$ and $M$ are symmetrically located on the opposite faces parallel to the $x$-$y$ plane (see figure). $V_1$ and $V_2$ are the potential differences between $K$ and $M$ in strips $1$ and $2$, respectively. Then, for a given current $I$ flowing through them in a given magnetic field strength $B$, the correct statement$(s)$ is(are):
$(A)$ If $w_1=w_2$ and $d_1=2d_2$, then $V_2=2V_1$
$(B)$ If $w_1=w_2$ and $d_1=2d_2$, then $V_2=V_1$
$(C)$ If $w_1=2w_2$ and $d_1=d_2$, then $V_2=2V_1$
$(D)$ If $w_1=2w_2$ and $d_1=d_2$, then $V_2=V_1$
$2.$ Consider two different metallic strips ($1$ and $2$) of same dimensions (lengths $\ell$, width $w$, and thickness $d$) with carrier densities $n_1$ and $n_2$, respectively. Strip $1$ is placed in magnetic field $B_1$ and strip $2$ is placed in magnetic field $B_2$, both along positive $y$-directions. Then $V_1$ and $V_2$ are the potential differences developed between $K$ and $M$ in strips $1$ and $2$, respectively. Assuming that the current $I$ is the same for both the strips, the correct option$(s)$ is(are):
$(A)$ If $B_1=B_2$ and $n_1=2n_2$, then $V_2=2V_1$
$(B)$ If $B_1=B_2$ and $n_1=2n_2$, then $V_2=V_1$
$(C)$ If $B_1=2B_2$ and $n_1=n_2$, then $V_2=0.5V_1$
$(D)$ If $B_1=2B_2$ and $n_1=n_2$, then $V_2=V_1$
Give the answer for question $1$ and $2$.

Two straight long conductors $AOB$ and $COD$ are perpendicular to each other and carry currents $i_1$ and $i_2$. The magnitude of the magnetic induction at a point $P$ at a distance $a$ from the point $O$ in a direction perpendicular to the plane $ACBD$ is