$A$ rectangular wire loop of sides $8 \text{ cm}$ and $3 \text{ cm}$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3 \text{ T}$ directed normal to the plane of the loop. The emf developed across the cut, if the velocity of the loop is $2 \text{ cm s}^{-1}$ in a direction normal to the shorter side of the loop, will be:

An aeroplane is travelling horizontally towards west with a speed of $540 \,km/h$. The wing span of the plane is $20 \,m$. If the horizontal component of the earth's magnetic field at the location is $2.5 \sqrt{3} \times 10^{-4} \,T$ and the dip angle is $30^{\circ}$, the potential difference developed between the ends of the wing is (in $\,V$)

One conducting $U$ tube can slide inside another as shown in the figure,maintaining electrical contacts between the tubes. The magnetic field $B$ is perpendicular to the plane of the figure. If each tube moves towards the other at a constant speed $v$,then the emf induced in the circuit in terms of $B, l$ and $v$,where $l$ is the width of each tube,will be

$A$ rectangular loop with a sliding connector of length $l = 1.0 \, m$ is situated in a uniform magnetic field $B = 2 \, T$ perpendicular to the plane of the loop. The resistance of the connector is $r = 2 \, \Omega$. Two resistors of $6 \, \Omega$ and $3 \, \Omega$ are connected as shown in the figure. The external force required to keep the connector moving with a constant velocity $v = 2 \, m/s$ is ........ $N$.

$A$ conducting circular loop is placed in a uniform magnetic field,$B = 0.025 \, T$,with its plane perpendicular to the field. The radius of the loop is made to shrink at a constant rate of $1 \, mm \, s^{-1}$. The induced $emf$ when the radius is $2 \, cm$ is:

$A$ 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 $\beta = 2$.