Motional EMF:- Due to Rotation of Conducting Rod E

MOTIONAL EMF: - DUE TO ROTATION OF CONDUCTING ROD EMF Induced due to translational motion of conducting rod in Magnetic field When a rod moves with a velocity v in a magnetic field B, the electrons in the rod feel a magnetic force pulling them downward. As a result, free electrons gather at the lower end, creating a shortage of electrons and an excess of positive charge at the upper end. This charge difference produces an electric field pointing downward, exerting an upward force on electrons. After the rod has been moving for a sufficient period, enough charges accumulate at the ends to balance the forces qE and qvB. Therefore, E = vB. [Diagram showing a conducting rod moving with velocity v in a magnetic field B (into the page). Positive charge accumulates at end P and negative charge at end Q, creating an electric field E. The magnetic force qvB on a charge is also shown.] V_P - V_Q = VBl The moving rod is equivalent to the following diagram, electrically. [Circuit diagram showing a voltage source B_l v in series with a resistor r, connected between points P and Q.] The illustration displays a closed coil ABCA in motion within a consistent magnetic field B at a velocity v. Since the flux passing through the coil remains constant, the induced electromotive force (emf) is zero. [Diagram of a rectangular coil ABCA moving with velocity v in a uniform magnetic field B (into the page).] Now, let's look at the rod AB, which is a segment of the coil. The induced emf in the rod is given by B l v. Let's assume that the induced emf in the ACB segment is denoted as E, as illustrated. [Diagram of the segment ACB of the coil, showing an induced emf vBL.] Since the emf in the coil is zero, Emf (in ACB) + Emf (in BA) = 0 Or -E + vBL = 0 Or E = vBL Thus emf induced in any path joining A and B is same, provided the magnetic field is uniform. Also the equivalent emf between A and B is BLV (here the two emf’s are in parallel) Ex. A rod of length l is placed perpendicular to a long wire carrying current I. The rod is moved parallel to the wire with a velocity v. Find the emf induced in the rod , if its nearest end is at a distance ‘a’ from the wire. [Diagram showing a conducting rod of length l moving with velocity v parallel to a wire carrying current I. The nearest end of the rod is at a distance 'a' from the wire. The magnetic field B is also indicated.] Sol. Consider a segment of rod of length dx, at a distance x, from the wire. Emf induced in the segment dE = (μ₀I / 2πx) dx. v ∴ E = ∫(from a to a+l) (μ₀I v dx) / (2πx) = (μ₀I v) / (2π) [ln x] (from a to a+l) = (μ₀I v) / (2π) ln((l+a)/a) Ex. A rectangular loop is moving parallel to a long wire carrying current I with a velocity v. Find the emf induced in the loop , if its nearest end is at a distance ‘a’ from the wire. Draw equivalent electrical diagram. [Diagram showing a rectangular loop of sides b and l moving with velocity v parallel to a wire carrying current I. The nearest side is at a distance 'a' from the wire.] Sol. emf = 0 ; [Equivalent circuit diagram showing two opposing voltage sources, each with magnitude (μ₀Iv / 2π) ln((a+b)/a), connected by resistors.] EMF induced due to rotational motion of conduction rod in magnetic field Rotation of the Rod Consider a conducting rod of length l rotating in a uniform magnetic field. [Diagram showing a rod of length l rotating with angular velocity ω in a uniform magnetic field B (into the page). A small segment dr is considered.] [Diagram showing the small segment dr behaving like a rod moving with velocity v = rω.] [Text box: v = rω, Small Segment Just like a rod] [Text box: dε = rω B dr, emf induced in a small segment] Emf induced in a small segment of length dr located at a distance r from the end = v B dr = rω B dr ∴ emf induced in the rod = ωB ∫(from 0 to l) r dr = 1/2 Bωl² equivalent of this rod is as following or e = dφ/dt e = (flux through the area swept by the rod in time dt) / dt e = (B(1/2 l² dθ))/dt = 1/2 Bl²ω [Circuit diagram showing a voltage source of Bωl²/2 in series with a resistor r.] Ex. A rod PQ of length 2l is rotating about one end P in a uniform magnetic field B which is perpendicular to the plane of rotation of the rod. Point M is the mid point of the rod. Find the induced emf between M & Q if that between P & Q is 100V. [Diagram of a rod PQ of length 2l rotating about end P in a uniform magnetic field B. Point M is the midpoint.] Sol. E_MQ = E_PQ - E_PM E_PM = (Bωl²)/2 = 100 E_PQ = (Bω(2l)²)/2 = (Bω4l²)/2 ⇒ E_MQ = (3/2) Bωl² = (3/2) x 100 V = 75 V 26.1s