The network shown in the figure is a part of a complete circuit. If at a certain instant the current $i$ is $5 \, A$ and is decreasing at the rate of $10^3 \, A/s$,then $V_A - V_B$ is......$V$.

$A$ square loop of side $12 \; cm$ with its sides parallel to $X$ and $Y$ axes is moved with a velocity of $8 \; cm \, s^{-1}$ in the positive $x$-direction in an environment containing a magnetic field in the positive $z$-direction. The field is neither uniform in space nor constant in time. It has a gradient of $10^{-3} \; T \, cm^{-1}$ along the negative $x$-direction (that is, it increases by $10^{-3} \; T \, cm^{-1}$ as one moves in the negative $x$-direction), and it is decreasing in time at the rate of $10^{-3} \; T \, s^{-1}$. Determine the direction and magnitude of the induced current in the loop if its resistance is $4.50 \; m\Omega$.

Two different coils have self-inductance $L_1 = 8 \, mH$ and $L_2 = 2 \, mH$. The current in both coils is increased at the same constant rate. At a certain instant of time,the power supplied to the two coils is the same. At that time,the current,induced voltage,and energy stored in the first coil are $i_1, V_1$,and $W_1$ respectively. Corresponding values for the second coil at the same instant are $i_2, V_2$,and $W_2$ respectively. Then:

In the $LR$ circuit shown,what is the variation of the current $I$ as a function of time? The switch is closed at time $t = 0 \, s.$

$A$ square coil $ABCD$ lies in the $x-y$ plane with its centre at the origin. $A$ long straight wire passing through the origin carries a current $i = 2t$ in the negative $z$-direction. The induced current in the coil is

$A$ square loop of side $1 \, m$ is placed in a perpendicular magnetic field. Half of the area of the loop is inside the magnetic field. $A$ battery of $emf$ $10 \, V$ and negligible internal resistance is connected in the loop. The magnetic field changes with time according to the relation $B = (0.01 - 2t) \, Tesla$. The resultant $emf$ in the loop will be.....$V$