Q.A cylindrical bar magnet is rotated about its axis. A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then

A rotating conducting cylindrical magnet is a homopolar (Faraday) generator.

• The bar magnet is itself a conductor. Its own axial field $\vec B$ threads the metal.
• As the cylinder spins with angular speed $\omega$, every free charge at radius $r$ moves with speed $v=\omega r$ across the axial field, so it feels a radial Lorentz force $q\,(\vec v\times\vec B)$. This pushes charge steadily between the axis and the rim.
• The wire (axis contact to a rim contact) closes the circuit, so a constant, one-directional (DC) current flows through it — its sense does not reverse, because the geometry of the field and the motion is unchanging in time.

Note the symmetry argument "field unchanged $\Rightarrow$ no emf" is a trap: it is a motional emf inside the moving conductor, not a flux change through a fixed loop.

The setup

A cylindrical bar magnet, which is a conductor, spins about its own geometric axis. One sliding contact sits on the axis and another on the curved surface; a wire joins them through an ammeter, completing a circuit.

Why a current flows — the motional-emf picture

The magnet's field is (roughly) axial, $\vec B \parallel$ axis, and it is carried rigidly with the metal. Consider a free electron in the metal at radius $r$ from the axis. Because the body rotates at angular speed $\omega$, that charge moves with velocity

i.e. tangentially. It therefore experiences a Lorentz force

With $\vec v$ tangential and $\vec B$ axial, $\vec v \times \vec B$ points radially. So charge is driven along the radius, between the axis and the rim. Integrating this force per unit charge from axis ($r=0$) to rim ($r=R$) gives a motional emf

which is constant in time. A constant emf drives a steady (DC) current $I=\varepsilon/R_\text{circuit}$ round the wire.

Why the "symmetry $\Rightarrow$ no emf" argument fails

It is tempting to say: a cylinder is symmetric about its axis, so rotating it leaves $\vec B$ unchanged everywhere, the flux through the circuit never changes, and hence there is no emf. That reasoning applies to a stationary loop linking a changing flux. Here the emf is motional — it lives in the moving conductor itself, where the flux rule is not the right tool. The charges genuinely move through the field, so a current genuinely flows.

Method: Recognising a Motional-EMF (Homopolar Generator) Problem in Disguise

Some induction questions describe a setup where Faraday's flux-rule instinct ("nothing changes, so no current") gives the WRONG answer — because the emf here is motional, generated inside a moving conductor itself, not by a changing flux through a fixed external loop.

Steps

Step 1: Check whether the conductor itself is moving through a field it's rigidly carrying along

If a conducting body is rotating (or translating) with a magnetic field that is fixed relative to the body itself (like a magnetized rotating cylinder), the standard "flux through a stationary loop" argument doesn't apply — the charges inside the moving conductor are the ones experiencing a force.

Step 2: Apply the Lorentz force to a free charge inside the moving conductor

Every free charge $q$ at some point in the moving conductor, with local velocity $\vec v$, feels

Work out the direction of $\vec v$ (e.g. tangential, for a point rotating about an axis) and the direction of $\vec B$ (e.g. along the axis) to find which way this force pushes charge — this is the origin of the emf, not a changing external flux.

Step 3: Integrate the force per unit charge along the conductor to get the emf

For a rigid body rotating at constant angular speed with a fixed field geometry, every term inside this integral is constant in time, so $\varepsilon$ itself comes out constant — this is the key signature of a steady (DC) motional emf, as opposed to the sinusoidally-varying emf you'd get from a coil rotating relative to an external field.

Step 4: Recognise WHY the naive symmetry argument fails

The trap in this class of question is reasoning "the setup looks unchanged from the outside (same field pattern, same geometry) at every instant, so nothing should happen." That test is only valid for a stationary loop linking an external, changing flux (Faraday's rule). Here the charges are physically moving through the field inside the conductor — a genuinely different physical mechanism (the Lorentz force) that produces a real, steady current regardless of the external symmetry.