An aeroplane having a wing span of $30 \ m$ flies due north with a speed of $170 \ m/s$. If the vertical component of the Earth's magnetic field is $B = 3.6 \times 10^{-5} \ T$, the potential difference between the tips of the wings will be: (in $mV$)

$A$ 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:

$A$ metallic disc of radius $0.3 \ m$ is rotating with a constant angular speed of $60 \ rad \ s^{-1}$ in a plane perpendicular to a uniform magnetic field of $5 \times 10^{-2} \ T$. The emf induced between a point on the rim and the centre of the disc is: (in $V$)

The loop shown moves with a velocity $v$ in a uniform magnetic field of magnitude $B$,directed into the paper. The potential difference between $P$ and $Q$ is $e$. Then:

$A$ rectangular loop with a sliding connector of length $10\, cm$ is situated in a uniform magnetic field perpendicular to the plane of the loop. The magnetic induction is $0.1\, T$ and the resistance of the connector is $1\, \Omega$. The sides $AB$ and $CD$ have resistances $2\, \Omega$ and $3\, \Omega$ respectively. Find the current in the connector during its motion with a constant velocity of $1\, m/s$.

$A$ long metal bar of $30\,cm$ length is aligned along a north-south line and moves eastward at a speed of $10\,ms^{-1}$. $A$ uniform magnetic field of $4.0\,T$ points vertically downwards. If the south end of the bar has a potential of $0\,V$,the induced potential at the north end of the bar is.....$V$.