$A$ closed loop is moving with constant velocity in a uniform magnetic field as shown. The potential difference and current in element $AB$ will be:

$A$ metallic conductor of length $1 \; m$ rotates in a vertical plane parallel to the east-west direction about one of its ends with 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 emf induced between the two ends of the conductor is .............

An aeroplane,with its wings spread $10 \, m$,is flying at a speed of $180 \, km/h$ in a horizontal direction. The total intensity of the Earth's magnetic field at that location is $2.5 \times 10^{-4} \, Wb/m^2$ and the angle of dip is $60^{\circ}$. The $EMF$ induced between the tips of the plane's wings will be ...... $mV$.

$A$ 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 mass $m$ and length $l$ moves on two frictionless parallel rails in the presence of a constant uniform magnetic field of magnitude $B$ directed into the page as shown in the figure. The bar is given an initial velocity $v_{0}$ towards the right at $t=0$. Then,the:

The loop shown moves with a constant velocity $V$ in a uniform magnetic field of magnetic induction $B$ directed into the paper. The potential difference between $P$ and $Q$ is: