$A$ copper rod $AB$ of length $l$ is rotated about end $A$ with a constant angular velocity $\omega$. The electric field at a distance $x$ from the axis of rotation is

$A$ square loop of area $25\,cm^2$ has a resistance of $10\,\Omega$. The loop is placed in a uniform magnetic field of magnitude $40.0\,T$. 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 $1.0\,s$ will be $..........\times 10^{-3}\,J$.

$(a)$ 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.

The figure shows an apparatus suggested by Faraday to generate electric current from a flowing river. Two identical conducting plates of length $a$ and width $b$ are placed parallel facing one another on opposite sides of the river flowing with velocity $u$ at a distance $d$ apart. Now both the plates are connected by a load resistance $R$. Then the current through the load $R$ is: (Consider the vertical component of the magnetic field produced by the earth is $B_v$ and the resistivity of river water is $\rho$.)

$A$ rod of $10 \ cm$ length is moving perpendicular to a uniform magnetic field of intensity $5 \times 10^{-4} \ Wb/m^2$. If the acceleration of the rod is $5 \ m/s^2$, then the rate of increase of induced $emf$ is . . . . . . .

$A$ wire of length $L$ is bent into a semicircle and moved with a velocity $v$ in a magnetic field $B$ as shown. What is the induced $emf$ between its two ends?