Q.A straight conducting rod of length $l$ slides with a uniform velocity $v$ along two parallel rails, in a region of uniform magnetic field $B$ perpendicular to the plane of the rails, with $v$, $B$ and the rod mutually perpendicular. Derive an expression for the motional emf induced between the ends of the rod, starting from the magnetic force on a free charge inside it.

Each spoke is an independent conducting rod rotating about the same axle in the same field, so each develops the SAME emf; connecting more of them in parallel between axle and rim does not add up their emfs, so E stays unchanged.

For a single conducting rod of length R rotating with angular speed omega in a field B (perpendicular to the plane of rotation), the motional emf between the centre and the rim is E = (1/2) B omega R^2. Every spoke, being identical in length and rotating at the same rate in the same field, develops this same emf E between the axle and the rim. All spokes are connected between the same two points (axle and rim), i.e. they are in parallel, not in series - so adding more spokes just adds more parallel paths of the same emf E, it does not increase the emf between the two terminals. Doubling the number of spokes therefore leaves E unchanged.

A rod rotating about one end sweeps out a circle; summing the motional emf $Bv\,dr$ over its length gives $\varepsilon = \tfrac12 B\omega l^2$.

Consider a small element of the rod at distance $r$ from the pivoted end, of length $dr$. Its linear speed is $v = \omega r$ (perpendicular to the rod, in the plane of rotation, hence also perpendicular to $B$). The motional emf induced across this element is:

$d\varepsilon = Bv\,dr = B\omega r\,dr$

Integrating from $r=0$ (pivot) to $r=l$ (free end), since all elemental emfs add up along the rod:

$\varepsilon = \int_0^l B\omega r\,dr = B\omega\left[\frac{r^2}{2}\right]_0^l = \frac{1}{2}B\omega l^2$

Motional emf ε = Bvl is a voltage; its unit is volt, option (vi).

When a conductor of length l moves with velocity v perpendicular to a magnetic field B, an emf is induced across it: ε = Bvl. This is an electromotive force, so its SI unit is the volt. Among the listed choices, the correct match is (vi) Volt.

Motional EMF in a moving conductor comes from the magnetic force on its free charges, which depends on v x B; if v is parallel to B this cross product vanishes.

The motional EMF induced in a straight conductor of length l moving with velocity v in field B is:

emf = (v x B) . l = B v l sin(phi)

where phi is the angle between v and B. Here the conductor moves parallel to the magnetic field, so phi = 0 degrees, and sin(0) = 0.

emf = B v l sin(0) = 0

Physically, the magnetic force on the free charges, F = q(v x B), is zero because v and B point in the same (or exactly opposite) direction, so there is no sideways push to separate charge along the conductor.

The induced emf in a moving conductor in a uniform magnetic field is purely a motional emf given by $\mathcal{E} = B l v$, which depends only on the magnetic field $B$, the length $l$ of the moving arm, and its velocity $v$. The resistance $r$ of the arm does not appear in this expression — it only determines the current that flows. Hence the correct answer is (c).

Fleming's right-hand rule gives the direction of INDUCED current (a motional-EMF/generator situation); the left-hand rule instead gives the direction of FORCE on a current-carrying conductor (a motor situation).

Orient the thumb, forefinger and middle finger of the right hand mutually perpendicular: thumb points along the direction of motion of the conductor, forefinger along the magnetic field, and the middle finger then gives the direction of the induced current.

The induced (motional) emf across the ends of the rod is $\varepsilon = Bvl$.

When a conducting rod of length $l$ moves with velocity $v$ perpendicular to a uniform magnetic field $B$ (with $B$, $v$ and the rod's length mutually perpendicular), each free charge experiences a magnetic force $qvB$. This separates charge until an electric field balances it, producing an emf.

The rod sweeps area at rate $\dfrac{dA}{dt} = lv$, so by Faraday's law

$\varepsilon = B\frac{dA}{dt} = Bvl.$

When a rod moves parallel to a magnetic field, the velocity and field are aligned, so the magnetic flux through any loop remains constant and no emf is induced. The answer is zero.