$A$ closed loop $PQRS$ carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments $PS, SR$ and $RQ$ are $F_1, F_2$ and $F_3$ respectively and are in the plane of the paper and along the directions shown,the force on the segment $QP$ is

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 $2$.

If two protons are moving with speed $v = 4.5 \times 10^{5} \, m/s$ parallel to each other,then the ratio of electrostatic and magnetic force between them is:

Match List-$I$ with List-$II$.

List-$I$	List-$II$
$(A)$ Permeability of free space	$(I) \ [M L^2 T^{-2}]$
$(B)$ Magnetic field	$(II) \ [M T^{-2} A^{-1}]$
$(C)$ Magnetic moment	$(III) \ [M L T^{-2} A^{-2}]$
$(D)$ Torsional constant	$(IV) \ [L^2 A]$

Choose the correct answer from the options given below:

An $\alpha$ particle of energy $10 \ eV$ is moving in a circular path in a uniform magnetic field. The energy of a proton moving in the same path and the same magnetic field will be [mass of $\alpha$ particle $= 4 \times$ mass of proton]. (in $eV$)

$A$ charge $Q$ is uniformly distributed over the surface of a nonconducting disc of radius $R$. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity $\omega$. As a result of this rotation,a magnetic field of induction $B$ is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity constant and vary the radius of the disc,then the variation of the magnetic induction at the centre of the disc will be represented by which of the following figures?