$A$ rod of length $80 \ cm$ rotates about its midpoint with a frequency of $10 \ rev/s$. The potential difference (in volts) between the two ends of the rod due to a magnetic field $B = 0.5 \ T$ directed perpendicular to the rod is:

$A$ wire $ab$ of length $l$,mass $m$,and resistance $R$ slides on a smooth,thick pair of metallic rails joined at the bottom as shown in the figure. The plane of the rails makes an angle $\theta$ with the horizontal. $A$ vertical magnetic field $B$ exists in the region. If the wire slides on the rails at a constant speed $v$,then which of the following can be correct for $B$?

$A$ square loop of area $25\,cm^2$ has a resistance of $10\,\Omega$. The loop is placed in a uniform magnetic field of magnitude $40.0\,T$. The plane of the loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in $1.0\,s$ will be $..........\times 10^{-3}\,J$.

$A$ 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 $1\, m/s$.

$A$ conducting wire is moving towards the right in a magnetic field $B$. The direction of the induced current $i$ in the wire is shown in the figure. The direction of the magnetic field will be

$A$ current carrying circular loop is perpendicular to a magnetic field of induction $10^{-4} \, T$. If the radius of the loop starts shrinking at a uniform rate of $2 \, mm/s$, then the emf induced in the loop at the instant, when its radius is $20 \, cm$ will be (in $\pi \, \mu V$)