$A$ player with a $3 \text{ m}$ long iron rod runs towards the east with a speed of $30 \text{ km/hr}$. The horizontal component of the Earth's magnetic field is $4 \times 10^{-5} \text{ Wb/m}^2$. If the player is running with the rod in horizontal and vertical positions,then the potential difference induced between the two ends of the rod in the two cases will be:

$A$ straight conductor of length $0.4 \ m$ is moving with a speed of $7 \ ms^{-1}$ perpendicular to a magnetic field of intensity $0.9 \ Wb \ m^{-2}$. The induced emf across the conductor will be (in $V$)

$A$ conducting square frame of side $a$ and a long straight wire carrying current $I$ are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity $V$. The $emf$ induced in the frame will be proportional to

$A$ 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

Derive the equation for the induced $emf$ in a rod of length $l$ sliding with velocity $v$ on a $U$-shaped frame placed perpendicular to a uniform magnetic field $B$.

$A$ rectangular conducting loop of length $4 \ cm$ and width $2 \ cm$ is in the $xy$-plane,as shown in the figure. It is being moved away from a thin and long conducting wire along the direction $\frac{\sqrt{3}}{2} \hat{x} + \frac{1}{2} \hat{y}$ with a constant speed $v$. The wire is carrying a steady current $I = 10 \ A$ in the positive $x$-direction. $A$ current of $10 \ \mu A$ flows through the loop when it is at a distance $d = 4 \ cm$ from the wire. If the resistance of the loop is $0.1 \ \Omega$,then the value of $v$ is. . . . . . $ms^{-1}$.
[Given: The permeability of free space $\mu_0 = 4 \pi \times 10^{-7} \ NA^{-2}$]