In the given branch $AB$ of a circuit,a current $I = (10t + 5) \, A$ is flowing,where $t$ is time in seconds. At $t = 0$,the potential difference between points $A$ and $B$ $(V_A - V_B)$ is.....$V$.

$A$ rectangular loop has a sliding connector $PQ$ of length $2\ m$ and resistance $10\Omega$. It is moving with a speed $5\ m/s$ as shown. The set-up is placed in a uniform magnetic field $3\ T$ directed into the plane of the paper. Find the three currents $I_1$,$I_2$,and $I$.

$A$ square loop of area $25 \, cm^2$ has a resistance of $10 \, \Omega$. The loop is placed in a uniform magnetic field of magnitude $40 \, T$. 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 $1 \, s$ will be:

Two identical conductors $P$ and $Q$ are placed on two frictionless rails $R$ and $S$ in a uniform magnetic field directed into the plane. If $P$ is moved in the direction shown in the figure with a constant speed,then rod $Q$

$A$ conducting rod of length $L$ rotates with angular speed $\omega$ in a uniform magnetic field of induction $B$ which is perpendicular to its motion. The induced emf developed between the two ends of the rod is

Consider a thin metallic sheet perpendicular to the plane of the paper moving with speed $v$ in a uniform magnetic field $B$ directed into the plane of the paper (See figure). If charge densities $\sigma_1$ and $\sigma_2$ are induced on the left and right surfaces,respectively,of the sheet,then (ignore fringe effects):