$A$ wheel with $10$ spokes,each of length $L \ m$,is rotated with a uniform angular velocity $\omega$ in a plane normal to a magnetic field $B$. What is the emf induced between the axle and the rim of the wheel?

$A$ 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 $\beta = 2$.

$A$ square loop $PQRS$ having $10$ turns, area $3.6 \times 10^{-3} \, m^2$ and resistance $100 \, \Omega$ is slowly and uniformly being pulled out of a uniform magnetic field of magnitude $B=0.5 \, T$ as shown. Work done in pulling the loop out of the field in $1.0 \, s$ is . . . . . $\times 10^{-6} \, J$.

$A$ horizontal telegraph wire of length $30 \ m$ spread east to west falls freely from a height of $20 \ m$. If the resistance of the wire is $40 \ \Omega$ and the horizontal component of the earth's magnetic field at the place is $2 \times 10^{-5} \ T$,then the induced current when the wire reaches the ground is (Acceleration due to gravity $= 10 \ m \ s^{-2}$)

$A$ rod of length $1.0 \,m$ is rotated in a plane perpendicular to a uniform magnetic field of induction $0.25 \,T$ with a frequency of $12 \,rev/s$. The induced emf across the ends of the rod is (in $\,V$)

$A$ rectangular loop has a sliding connector $PQ$ of length $l$ and resistance $R \ \Omega$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $I_1, I_2$ and $I$ are: