$A$ rod of length $l$ is oscillating as a physical pendulum about one of its ends with a small angular amplitude $\alpha$ in a crossed magnetic field $B$. The maximum $emf$ induced in the rod will be:

$A$ conducting bar of length $L$ is free to slide on two parallel conducting rails as shown in the figure. Two resistors $R_{1}$ and $R_{2}$ are connected across the ends of the rails. There is a uniform magnetic field $\vec{B}$ pointing into the page. An external agent pulls the bar to the left at a constant speed $v$. The correct statement about the directions of induced currents $I_{1}$ and $I_{2}$ flowing through $R_{1}$ and $R_{2}$ respectively is:

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.....$V$.

$A$ wheel with $10$ metallic spokes,each $0.5 \ m$ long,is rotated with a speed of $120 \ rev/min$ in a plane normal to the horizontal component of Earth's magnetic field $H_E$ at a place. If $H_E = 0.4 \ G$ at the place,then the induced emf is: $(1 \ G = 10^{-4} \ T)$

To measure a magnetic field between the magnetic poles of a loudspeaker, a small coil having $30$ turns and $2.5 \, cm^2$ area is placed perpendicular to the field and removed immediately. If the total charge flown through the coil is $7.5 \times 10^{-3} \, C$ and the total resistance of the wire and galvanometer is $0.3 \, \Omega$, then the magnitude of the magnetic field is

$A$ metal rod of length $1\,m$ is rotated about one of its ends in a plane at right angles to a magnetic field of induction $2.5 \times 10^{-3}\,Wb/m^2$. If it makes $1800\,rpm$,calculate the induced e.m.f. between its ends in $V$.