$A$ metallic rod of length $1 \,m$ held along the east-west direction is allowed to fall down freely. Given the horizontal component of the Earth's magnetic field $B_H = 3 \times 10^{-5} \,T$. The emf induced in the rod at an instant $t = 2 \,s$ after it is released is (Take $g = 10 \,m/s^2$):

$A$ conductor of length $l$ and mass $m$ can slide along a pair of vertical metal guides connected by a resistance $R$,as shown in the figure. Friction,resistance of the conductor,and guide rails are negligible. There exists a horizontal uniform magnetic field of strength $B$ normal to the plane of the page and directed outward. The terminal speed of fall under the influence of gravity is:

$A$ $1.0 \; m$ long metallic rod is rotated with an angular frequency of $400 \; rad \; s^{-1}$ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. $A$ constant and uniform magnetic field of $0.5 \; T$ parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring. (in $; V$)

$A$ horizontal straight wire $10 \; m$ long extending from east to west is falling with a speed of $5.0 \; m \, s^{-1}$,at right angles to the horizontal component of the earth's magnetic field,$0.30 \times 10^{-4} \; Wb \, m^{-2}$.
$(a)$ What is the instantaneous value of the $emf$ induced in the wire?
$(b)$ What is the direction of the $emf$?
$(c)$ Which end of the wire is at the higher electrical potential?

$A$ rod of length $1.0 \,m$ is rotated in a plane perpendicular to a uniform magnetic field of induction $0.25 \,T$ with a frequency of $12 \,rev/s$. The induced emf across the ends of the rod is (in $\,V$)

$A$ thin wire of length $2 \ m$ is perpendicular to the $xy$-plane. It moves with a velocity $v = (2\hat{i} + 3\hat{j} + \hat{k}) \ m/s$ in a magnetic field $B = (\hat{i} + 2\hat{j}) \ Wb/m^2$. What is the induced potential difference (emf) across the ends of the wire?