$A$ straight conductor $0.1 \ m$ long moves in a uniform magnetic field of $0.1 \ T$. The velocity of the conductor is $15 \ m/s$ and is directed perpendicular to the field. The emf induced between the two ends of the conductor is: (in $V$)

$A$ metal conductor of length $1\;m$ rotates vertically about one of its ends at an angular velocity of $5\;rad/s$. If the horizontal component of the Earth's magnetic field is $0.2 \times 10^{-4}\;T$,then the $e.m.f.$ developed between the two ends of the conductor is:

$(a)$ 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.

$A$ straight conductor of length $4 \,m$ moves at a speed of $10 \,m/s$. When the conductor makes an angle of $30^{\circ}$ with the direction of a magnetic field of induction $0.1 \,Wb/m^2$, the induced emf is: (in $\,V$)

$A$ conducting bar moves on two conducting rails as shown in the figure. $A$ constant magnetic field $B$ exists into the page. The bar starts to move from the vertex at time $t=0$ with a constant velocity $v$. If the induced $\text{EMF}$ is $E \propto t^n$,then the value of $n$ is . . . . . . .

An infinitely long straight wire carrying current $I$,an open rectangular loop,and a conductor $C$ with a sliding connector are located in the same plane,as shown in the figure. The connector has length $l$ and resistance $R$. It slides to the right with a velocity $v$. The resistance of the conductor and the self-inductance of the loop are negligible. The induced current in the loop,as a function of separation $r$ between the connector and the straight wire,is: