Consider the situation shown in the figure. The wire $AB$ is sliding on the fixed rails with a constant velocity. If the wire $AB$ is replaced by a semicircular wire,the magnitude of the induced current will

$A$ $10\,m$ long horizontal wire extends from North-East to South-West. It is falling with a speed of $5.0\,m/s$,at right angles to the horizontal component of the earth's magnetic field,which is $0.3 \times 10^{-4}\,Wb/m^2$. The value of the induced $emf$ in the wire is:

One conducting $U$ tube can slide inside another as shown in the figure,maintaining electrical contacts between the tubes. The magnetic field $B$ is perpendicular to the plane of the figure. If each tube moves towards the other at a constant speed $v$,then the emf induced in the circuit in terms of $B, l$ and $v$,where $l$ is the width of each tube,will be

$A$ conducting rod $AB$ of length $l$ is projected on a frictionless frame $PSRQ$ with velocity $v_0$ in a uniform magnetic field $\vec{B}$ directed into the plane. The velocity of the rod after time $t$ is

$A$ metal wire $PQ$ slides on parallel metallic rails having separation $0.25 \ m$,each having negligible resistance. There is a $2 \ \Omega$ resistor and $10 \ V$ battery as shown in the figure. There is a uniform magnetic field directed into the plane of the paper of magnitude $0.5 \ T$. $A$ force of $0.5 \ N$ to the left is required to keep the wire $PQ$ moving with constant speed to the right. With what speed is the wire $PQ$ moving? ..... $m/s$ (Neglect self-inductance of the loop)

$A$ conducting rod $AC$ of length $4l$ is rotated about a point $O$ in a uniform magnetic field $\vec B$ directed into the paper. $AO = l$ and $OC = 3l$. Then: