$A$ wire loop enclosing a semi-circle of radius $R$ is located on the boundary of a uniform magnetic field of induction $\vec{B}$. At time $t=0$,the loop is set into rotation with angular velocity $\omega$ about its axis $0$,coinciding with a line of vector $\vec{B}$ on the boundary as shown in the figure. The emf induced in the loop is

$A$ copper rod $AB$ of length $l$ is rotated about end $A$ with a constant angular velocity $\omega$. The electric field at a distance $x$ from the axis of rotation is

$A$ conducting wire $XY$ of mass $m$ and negligible resistance slides smoothly on two parallel conducting wires as shown in the figure. The closed circuit has a resistance $R$ due to $AC$. $AB$ and $CD$ are perfect conductors. There is a magnetic field $\vec{B} = B(t) \hat{k}$.
$(i)$ Write down the equation for the acceleration of the wire $XY$.
$(ii)$ If $\vec{B}$ is independent of time,obtain $v(t)$,assuming $v(0) = u_0$.
$(iii)$ For $(ii)$,show that the decrease in kinetic energy of $XY$ equals the heat lost in $R$.

$A$ bar of mass $m$,length $d$,and resistance $R$ slides without friction in a horizontal plane,moving on parallel rails as shown in the figure. $A$ battery that maintains a constant emf $\varepsilon$ is connected between the rails,and a constant magnetic field $\vec{B}$ is directed perpendicularly to the plane of the page. Assuming the bar starts from rest,find the speed at time $t$.

$A$ rectangular loop circuit has a sliding wire $PQ$ as shown in the figure. The loop is placed in a magnetic field $B$,perpendicular to its plane. The resistance of the wire $PQ$ is $R$. If the wire moves with constant velocity $v$,then find the current flowing in the wire $PQ$?

$A$ conducting wire of parabolic shape,initially $y=x^2$,is moving with velocity $\vec{V} = V_0 \hat{i}$ in a non-uniform magnetic field $\vec{B} = B_0 \left(1 + \left(\frac{y}{L}\right)^\beta\right) \hat{k}$,as shown in the figure. If $V_0, B_0, L$ and $\beta$ are positive constants and $\Delta \phi$ is the potential difference developed between the ends of the wire,then the correct statement$(s)$ is/are:
$(1)$ $|\Delta \phi|$ remains the same if the parabolic wire is replaced by a straight wire,$y=x$ initially,of length $\sqrt{2} L$.
$(2)$ $|\Delta \phi|$ is proportional to the length of the wire projected on the $y$-axis.
$(3)$ $|\Delta \phi| = \frac{1}{2} B_0 V_0 L$ for $\beta = 0$.
$(4)$ $|\Delta \phi| = \frac{4}{3} B_0 V_0 L$ for $\beta = 2$.