As shown in the figure,a metal rod makes contact with a partial circuit and completes the circuit. The circuit area is perpendicular to a magnetic field with $B = 0.15\, T$. If the resistance of the total circuit is $3\,\Omega$,the force needed to move the rod as indicated with a constant speed of $2\, ms^{-1}$ will be equal to:

Refer to the figure. The arm $PQ$ of the rectangular conductor is moved from $x=0$ outwards. The uniform magnetic field is perpendicular to the plane and extends from $x=0$ to $x=b$ and is zero for $x>b$. Only the arm $PQ$ possesses substantial resistance $r$. Consider the situation when the arm $PQ$ is pulled outwards from $x=0$ to $x=2b$ and is then moved back to $x=0$ with constant speed $v$. Obtain expressions for the flux,the induced emf,the force necessary to pull the arm,and the power dissipated as Joule heat. Sketch the variation of these quantities with distance.

$A$ coil has $1000$ turns and $500 \text{ cm}^2$ as its area. The plane of the coil is placed at right angles to a magnetic induction field of $2 \times 10^{-5} \text{ Wb/m}^2$. The coil is rotated through $180^{\circ}$ in $0.2 \text{ s}$. The average emf induced in the coil,in $\text{mV}$,is

$A$ 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?

What is the cause of the force acting on a sliding conductor placed in a magnetic field?

An equilateral triangular loop $ADC$ moves out of a finite magnetic field $B$ as shown in the figure. At time $t = 0$, side $DC$ of the loop is at the edge of the magnetic field. The magnetic field is perpendicular to the paper inwards (or perpendicular to the plane of the coil). The induced current versus time graph will be as: