An aeroplane in which the distance between the tips of wings is $50 \ m$ is flying horizontally with a speed of $360 \ km/hr$ over a place where the vertical component of the Earth's magnetic field is $2.0 \times 10^{-4} \ Wb/m^2$. The potential difference between the tips of the wings would be $V$.

$A$ conducting rod of length $l$ is falling with a velocity $v$ perpendicular to a uniform horizontal magnetic field $B$. The potential difference between its two ends will be

$A$ conducting rod $AB$ of length $l = 1\,m$ is moving with a velocity $v = 4\,m/s$. The velocity vector makes an angle of $30^o$ with the length of the rod. $A$ uniform magnetic field $B = 2\,T$ exists in a direction perpendicular to the plane of motion. Then:

$A$ conducting bar of length $L$ is free to slide on two parallel conducting rails as shown in the figure. Two resistors $R_{1}$ and $R_{2}$ are connected across the ends of the rails. There is a uniform magnetic field $\vec{B}$ pointing into the page. An external agent pulls the bar to the left at a constant speed $v$. The correct statement about the directions of induced currents $I_{1}$ and $I_{2}$ flowing through $R_{1}$ and $R_{2}$ respectively is:

The direction of current induced in a wire moving in a magnetic field is found using

$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with a 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: