MORE INFORMATION ON ELECTROMAGNETIC FIELDS AND WAVES

The Battery must do work to keep the potential difference b/w the terminals constant when a current flows in a wire joining the terminals. The work done moving a charge q from A to B terminals is Vq = q ∫ABE.dl Vq = ∫ABF.dl Where F is the force on the charge q (the electron) V = 1/q∫ABF.dl This also represents the e.m.f. of the battery With A and B as terminals. Note that an e.m.f. can exist in a closed circuit (loop) as well as across the terminals of a battery. Whenever the magnetic flux through a closed circuit is changing, the line integral of the force on a charge, integrated around the whole circuit is not zero. In this case, work must be done to take a charge around the circuit, and the work done may be used to define an e.m.f. just as the e.m.f. of the battery explained earlier on. The e.m.f., V around the circuit is given by V = 1/q∮ABF.dl If the closed circuit consists of a conductor of resistance R and contains no other source of e.m.f., other than the one due to changing magnetic fields, then a current I = V/R will flow around the circuit. Motional electromotive force Consider a situation in which an e.m.f. exists in a simple closed circuit. Let us consider its value A metal rod moving with velocity v perpendicular to a uniform field B In the diagram, let us consider a metal rod moving with constant velocity v in a direction perpendicular to a uniform magnetic field B The motion of a charged particle eg an electron in a magnetic field is due to the force it experiences This is given by F = qv^B Magnetic forces always act perpendicularly to the direction of motion of the charged particles. The magnitude of the force is found to be proportional to the speed of the particle, its charge q, and to the sine of the angle between the vectors V and B hence F ∝ qv ^ B The unit of the magnetic field B are defined by choosing the constant of proportionality to be unity. F = qv ^ B This is the Lorentz force law The Lorentz force law has components. F = Fx + Fy + Fz Fx = q(VyBz - VzBy) Fy = q(VzBx - VxBz) Fz = q(VxBy - VyBx) As well as the forces exerted on charges by electric fields, there may be additional magnetic forces experienced by moving charged particles. If an electric field E is present as well as a magnetic field B, a charge q experiences an additional force qE and the total force is, F = qE + qv ^ B This equation defines the fields E and B in free space, where the force on a moving test charge can be measured. Note that the test charge must be small enough not to affect the charge and current distribution s that give rise to the fields, otherwise the measured force will not give those fields present without the test charge. {v ≡ is velocity} Back to Motional ElectroMotiveForce. The magnetic force given by F = qv ^ B This magnetic force on each electron and positive charge in the rod will cause the free electrons in the rod to move towards end a and collect there. F ∝ qv ^ B This gives rise to a charge distribution which produces an electric field which halts further migration of free electrons. The electric field set up produces a force –eE on the electrons exactly equal and opposite to the force -eV ^ B (ie qv ^ B ⇨ the charge q ≡ -e) Which acts on them because of their motion in the magnetic field. Hence -eE = ev ^ B Hence the electric field is E = -v ^ B The potential difference Vab between the ends of the rod and b is given by Vba = ∫baE.dl Hence Vab = vBL Where L is the length of the rod. When you integrate from a to b with respect to .dl you get the length L. Ie Vba = ∫baE.dl = ∫ba-v^B.dl = ∫ba-vxB.dl = ∫ba-|v|B||sinθ.dl = vBL Hence Vab = vBL Where L is the length of the rod. EXAMPLE AN AEROPLANE WITH A WING SPAN OF 50m FLYING HORIZONTALLY AT 800Km PER HOUR AT A PLACE WHERE VERTICAL COMPONENT OF THE EARTH’S MAGNETIC FIELD IS 2x 10-5 T (TESLA) USING EQN BELOW Vba = Vbl There is a voltage difference between thw wing tips Equals 0.22 volts. In the illustration using the metal rod, there is a potential difference between the ends of th rod but no current flow is produced. This is because this constitutes an open circuit. If a metal rod moves on stationary conducting rails, perpendicularly to a uniform magnetic field, a current will flow because a complete circuit has been formed. A metal rod moving with velocity v on stationary conducting rails. A uniform field B acts as shown. In this case, charge do not build up on the ends a and b rather electrons traverse the complete circuit from b to a and around to b again; and continues to circulate. The line integral of the force on a charge q evaluated around the complete circuit is given by ∮F.dl = qvBL L is the length of the rod. NB: Note that only rod ab is in motion with velocity v. Hence only charges or electrons on the rod ab are moving with v velocity. The sole contribution to the line integral comes from the portion ab of the loop. The e.m.f. in the closed circuit is given by 1/q∮F.dl = vBL This can be called a motional e.m.f. because it is produced by the motion of a conductor in a magnetic field. The motional e.m.f. causes a current I to flow given by I = (vBL)/R Where R is the resistance of the circuit. Power is given by I2R Therefore, The instantaneous Power is given by (vBL)2/R The work is derived from the mechanical force needed to move the rod along the rails. WE CAN EXTEND THIS CONCEPT A BIT FURTHER TO DEVELOP AN ALTERNATING CURRENT OR E.M.F. USING THE SAME PRINCIPLE. In this case, instead of a rod, we shall consider a rectangular circuit moving with constant speed v through a region of uniform magnetic field. Consider the diagram below. At first only the portion ab is within the field. This means that induced e.m.f. will cause current to flow as the electrons will migrate towards a from d round the circuit and back to b again. This will reperesent current flow in one direction. So long as this part of the conductor is the only part of the conductor at right angle to the direction of motion of the conductor v, it will be the only contributor to this emf and hence the current flow. Since sin (90)=1 and sin(0) = 0 The situation changes when the entire rectangular circuit is in the magnetic field. At this stage, the end cd comes into the picture. Again electrons migrate towards d from c as a result of the Lorentz force experienced by the charged particles, in this case, the free electrons. These electrons move from c to d and back to c again. Notice that this direction is opposite the direction of the electron motion described earlier on when only ab part of the rectangular circuit was in the magnetic field. The result of this is that the two effects will cancel out the the resulting electromagnetic force will be zero as well as the current flow. That is the current flow will also be zero. (Note that we assume all things being equal. Ie that the magnetic field is uniform everywhere withing the magnetic region and that the rectangular circuit is regular uniform and made of homogenous material. Ie the ends ba and cd are identical.) When the rectangular circuit moves so that the end ba is now outside the magnetic region or field, its effect ceases to be felt. We are now left with only the effect of the end cd which causes the electrons to migrate from c to d and back to c again. This brings about a current flow in a direction opposite the current flow we experienced at the beginning of the experiment. SUMMARY WHEN THE COIL IS ENTERING THE FIELD, THERE IS AN EMF OF MMAGNITUDE vBL IN THE CIRCUIT AND THIS PRODUCES A CURRENT AS SHOWN IN THE DIAGRAM WHENHEN THE WHOLE COIL IS MOVING WITHIN THE UNIFORM FIELD, THE E.M.F. AND CURRENT ARE ZERO. THIS IS BECAUSE THE MAGNETIC FORCES ALONG THE SIDES ab AND cd CANCEL EACH OTHER OUT. WHEN THE COIL LEAVES THE FIELD THERE IS AGAIN AN E.M.F. EQUAL TO vBL BUT IT NOW ACTS IN THE OPPOSITE DIRECTION. The diagram above shows the variation of current with time for the moving circuit shown in the figure earlier on. (the rectangular circuit) The current is not exactly as shown; it cannot rise and fall infinitely quickly. The e.m.f. in the coil as it enters or leaves the field can be expressed in terms of the total magnetic flux through the coil. This flux is the sum of the part BLx [B- flux density, Lx – cross sectional area] due to the external field, where x is the length of the coil in the field, and the part due to the induced current in the coil itself. Since the induced current and the area of the coil are constant, this later contribution to the magnetic flux is constant. And 𝜕Φ/𝜕t = BL𝜕x/𝜕t = BLv The magnitude of the emfis given by |e.m.f.| = 𝜕Φ/𝜕t H-magnetic field strength H- magnetic field intensity B=𝜇H B = 𝝻H B – magnetic field density B – Magnetic flux density B = 𝝻𝝻0H = 𝜇𝜇0H 𝝻 = 𝜇 = RELATIVE PERMEABILITY 𝜇0 = 𝝻0 = PERMEABILITY IN FREE SPACE OR IN A VACUUM RETARDED POTENTIALS