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

$A$ circular coil of $500$ turns encloses an area of $0.04 \,m^2$. $A$ uniform magnetic field of induction $0.25 \,Wb/m^2$ is applied perpendicular to the plane of the coil. The coil is rotated by $90^o$ in $0.1 \,s$ at a constant angular velocity about one of its diameters. $A$ galvanometer of resistance $25 \,\Omega$ is connected in series with the coil. The total charge that will pass through the galvanometer is.......$C$

$A$ 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:

$A$ circular disc of radius $0.1 \ m$ is rotating about its axis at a frequency of $10 \ rev/s$ in a uniform magnetic field of $0.1 \ T$ directed perpendicular to the disc. Calculate the induced $emf$ between the center and the rim of the disc.

$A$ rectangular loop has a sliding connector $PQ$ of length $l$ and resistance $R \ \Omega$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $I_1, I_2$ and $I$ are:

$A$ rod of length $20 \text{ cm}$ is rotating with an angular speed of $100 \text{ rps}$ in a magnetic field of strength $0.5 \text{ T}$ about one of its ends. What is the potential difference between the two ends of the rod? $(V)$