$A$ conductor wire $ABCDE$ with each arm $10 \ cm$ in length is placed in a magnetic field of $\frac{1}{\sqrt{2}} \ T$,perpendicular to its plane. When the conductor is pulled towards the right with a constant velocity of $10 \ cm/s$,the induced emf between points $A$ and $E$ is . . . . . . $mV$.

$(a)$ Obtain an expression for the mutual inductance between a long straight wire and a square loop of side $a$ as shown in Figure.
$(b)$ Now assume that the straight wire carries a current of $50\; A$ and the loop is moved to the right with a constant velocity, $v=10\; m / s$. Calculate the induced $emf$ in the loop at the instant when $x=0.2\; m$. Take $a=0.1\; m$ and assume that the loop has a large resistance.

$A$ metallic rod of length '$L$' is rotated with an angular speed of '$\omega$' normal to a uniform magnetic field '$B$' about an axis passing through one end of the rod,as shown in the figure. The induced emf will be:

$A$ metallic cube of side $15\,cm$ is moving along the $y$-axis at a uniform velocity of $2\,m/s$ in a region of uniform magnetic field of magnitude $0.5\,T$ directed along the $z$-axis. In equilibrium,the potential difference between the faces of higher and lower potential developed because of the motion through the field will be $..........mV$.

$A$ square loop of side $a$ and resistance $R$ is moved in a region of uniform magnetic field $B$ (the loop remaining completely inside the field) with a velocity $v$ through a distance $x$. The work done is:

$A$ 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: $(1 \ G = 10^{-4} \ T)$