$A$ wheel with $10$ spokes,each of length $L \ m$,is rotated with a uniform angular velocity $\omega$ in a plane normal to a magnetic field $B$. What is the emf induced between the axle and the rim of the wheel?

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

$A$ pair of parallel conducting rails lie at a right angle to a uniform magnetic field of $2.0\, T$ as shown in the figure. Two resistors,$10\,\Omega$ and $5\,\Omega$,slide without friction along the rails. The distance between the conducting rails is $0.1\, m$. Then:

$A$ frame $CDEF$ is placed in a region where a magnetic field $\vec{B}$ is present. $A$ rod $PQ$ of length $l = 1 \, m$ moves with a constant velocity $v = 20 \, m/s$ and the strength of the magnetic field is $B = 1 \, T$. The power spent in the process is .............. $kW$ (take $R = 0.2 \, \Omega$ and assume all other wires and the rod have zero resistance).

$A$ boat is moving due east in a region where the earth's magnetic field is $3.6 \times 10^{-5} \text{ T}$ due north and horizontal. The boat carries a vertical conducting rod $2 \text{ m}$ long. If the speed of the boat is $2.00 \text{ m/s}$, the magnitude of the induced e.m.f. in the rod is: (in $\text{ mV}$)

$A$ circular wheel with $10$ spokes, with its plane vertical along East-West, is rotating about its natural axis with a uniform speed of $100$ revolutions per minute in the Earth's magnetic field. The radius of the wheel is $0.3 \, m$. If the $EMF$ induced between the centre of the wheel and the rim is $3 \pi \mu V$, what is the angle of dip at that place? (Vertical component of the Earth's magnetic field $B_{V} = 15 \mu T$)