$A$ jet plane is travelling towards west at a speed of $1800\; km/h$. What is the voltage difference developed between the ends of the wing having a span of $25\; m$,if the Earth's magnetic field at the location has a magnitude of $5 \times 10^{-4}\; T$ and the dip angle is $30^{\circ}$ (in $; V$)?

$A$ square loop $PQRS$ having $10$ turns, area $3.6 \times 10^{-3} \, m^2$ and resistance $100 \, \Omega$ is slowly and uniformly being pulled out of a uniform magnetic field of magnitude $B=0.5 \, T$ as shown. Work done in pulling the loop out of the field in $1.0 \, s$ is . . . . . $\times 10^{-6} \, J$.

$A$ conducting rod of length $L$ lies in the $XY$-plane and makes an angle $30^{\circ}$ with the $X$-axis. One end of the rod is initially at the origin. $A$ magnetic field exists in the region pointing along the positive $Z$-direction. The magnitude of the magnetic field varies with $y$ as $B = B_0 \left(\frac{y}{L}\right)^3$,where $B_0$ is a constant. At some instant,the rod starts moving with a velocity $v_0$ along the $X$-axis. The emf induced in the rod is

$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with uniform speed $v = 20\, m/s$ in a uniform magnetic field $B = 4.0\, T$ directed into the paper. $A$ capacitor of capacity $C = 10\, \mu F$ is connected as shown in the figure. Then:

$A$ square loop of area $25 \, cm^2$ has a resistance of $10 \, \Omega$. The loop is placed in a uniform magnetic field of magnitude $40 \, T$. 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 $1 \, s$ will be:

$A$ $1 \ m$ long metal rod $AB$ completes the circuit as shown in the figure. The area of the circuit is perpendicular to the magnetic field of $0.10 \ T$. If the resistance of the total circuit is $2 \ \Omega$,then the force needed to move the rod towards the right with a constant speed $(v)$ of $1.5 \ m/s$ is . . . . . . $N$.