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Physics Mechanics Notes

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Published in: Physics
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Physics Notes- 1. Mechanics 2. Magnetism 3. Electromagnetic Induction

Pankaj J / Mumbai

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Qualification: B.Tech/B.E. (KU - 2008)

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  1. ELECTRO - MAGNETIC INDUCTION INTRODUCTION Whenever magnetic flux linked with a circuit changes, an e.m.f. is induced in the circuit. If the circuit is closed, a current is also induced in it. The e.m.f. and current produced lasts as long as the flux linked with the circuit changes. The phenomenon is called electromagnetic induction MAGNETIC FLUX f é.dS f é.dS GAUSS'S LAW FOR MAGNETISM fé.dS - o FARADAY'S LAW Faraday performed three experiment : Experiment 1 : He pulled a loop of wire to the right through a magnetic field as soon figure (a). A current flowed in the loop. Experiment 2: He moved the magnet to the left, holding the loop still as soon figure (b). Again, a current flowed in the loop. Experiment 3: With both the loop and the magnet at rest as soon figure (c), he change the strength of the field (he used an electromagnet, and varied the current in the coil). Once again current flowed in the loop. x x x x (a) x x x x (b) of course, changing (c) magnetic field is an example of motional emf, conveniently expressed by the flux The first experiment, rule: do dt Faraday had an ingenious inspiration: A changing magnetic field induces an electric field. It is this "induced" electric field that accounts for the emf in Experiment 2. Indeed, if (as Faraday found empirically) the emf is again equal to the rate of change of the flux, do = $E.dl dt
  2. then E is related to the change in B by the equation f ey.da fE.dl This is Faraday's law, in integral form. Example 1 : Solution : Example 2 : Solution : A coil is placed in a constant magnetic field . The magnetic field is parallel to the plane of the coil as shown in figure. Find the emf induced in the coil. = 0 (always) since area is perpendicular to magnetic field. . emf = 0 Figure shows a coil placed in decreasing magnetic field applied perpendicular to the plane of coil . The magnetic field is decreasing at a rate of 10T/s. Find out current in magnitude and direction A=2m2 R=5Q dB emf = A. = 2 x 10=20 v dt . i = 20/ 5 = 4 amp. From lenz's law direction of current will be anticlockwise. LENZ'S LAW The induced current will flow in such a direction that the flux it produces tends to cancel the change. (a) x x x x x x (b) x x x x x x (c) Refer figure (a) A loop abcd enters a uniform magnetic field B at constant speed v. Using Faraday's equation, I el dt d(BS) d(B/x) dx Blv dt dt dt For the direction of current, we can use Lenz's law. As the loop enters the field @ magnetic field passing through the loop increases, hence, current in the loop is anticlockwise. (O IN). From the theory of motional emf, e = Bvl and using right hand rule, current in the circuit is anticlockwise. Thus, we see that e and e = Bvl give the same result. In the similar manner we can show that dt current in the lop in figure (b) is zero and in figure (c) it is clockwise. INDUCED EMF IN A ROD e = Blvsin0
  3. Example 3 : Solution : INDUCTOR A rod PQ of length L moves with a uniform velocity v parallel to a long straight wire carrying a current l. The end P remains at a fixed perpendicular distance r from the wire. Calculate the e.m.f. induced across the rod. The rod moves in the magnetic field produced by the current carrying wire as a result of which an e.m.f. is induced across the rod. Let us consider a small element of length dx of the rod at a distance x and x + dx from the wire. The e.m.f. induced across the element, d: = Bvdx The magnetic field Bat a distance x from a wire carrying a current lis given by 21tx Using (2) in (1), we get d: — -Evdx 21tx Induced emf across the entire length of the rod PQ is given by —v dx —v dx 21tx Ptolv r +Ldx Violv [loge x] 27t 211 x polv —[loge(r +L) — loge r] 211 Ptolv loge 211 Already we have studied about capacitor — that stores energy in the form of electric field. Like capacitor, inductor is also quite a commonly used element in electric circuits, which stores magnetic energy. Inductance of an inductor depends on its geometry and medium in which it lies. As we know that when current flows through a conductor a magnetic field is set-up around it, and hence it is associated with magnetic flux. If magnetic flux associated with a coil is (l) and current in it is l, then its inductance is given by the expression L 'L' is called self-inductance of the coil. S.l. unit of inductance is henry. INDUCED EMF IN A ROTATING COIL e = NABosinot = eo sinot SELF INDUCTANCE dt di dt CALCULATION OF SELF-INDUCTANCES (a) A long straight solenoid : For a very long solenoid, I > > a and equation reduces to
  4. N2A L = Ttpon2a21 (b) (c) An open-air transmission line . POI b A coaxial cable: L —In POI d —a —In a MUTUAL INDUCATNCE dt 27t a di p dt CALCULATION OF MUTUAL INDUCTANCES (a) Two solenoids : M = gmn A n = number of turns per unit length of the primary solenoid m = number of turns per unit length of the secondary solenoid (b) Example 4 : Solution : Two parallel coaxial circular loops: TtktoNIN2a2b2 3 2(a2 + What is the self inductance of a system of co-axial cables carrying current in opposite directions as shown. Their radii are 'a' and 'b' respectively. The 'B' between the space of the cables is B = POI/2Ttr The Ampere's law tells that 'B' outside the cables is zero, as the net current through the amperian loop would be zero. Taking an element of length C and thickness 'dr', cid) through it is CIG) — —L Cdr (I) 21tr 27t r r 27t a b In 1 211 a R D SELF-INDUCTANCES IN SERIES AND PARALLEL (a) Series connection : L = I-I +1-2 +2M LIL2 -M2 (b) Parallel connection : L RELATION BETWEEN SELF AND MUTUAL INDUCTANCE : COEFFICIENT OF COUPLING k Thus for an ideal coupling, k = 1 or M And for non-ideal coupling, k < 1 or M < Al/ÜV2 Example 5 : Figure shows two concentric coplanar coils with radii a and b (a
  5. Solution : (a) (b (c) (a) (b) b a Find the mutual inductance of the two coils Find the emf induced in the larger coil If the resistance of the larger loop is R find the current in it as a function of time To find mutual inductance, it does not matter in which coil we consider current and in which flux is calculated (Reciprocity theorem) Let current i be flowing in the larger coil. Magnetic field at the centre = 2b = h—Tta 2 flux through the smaller coil 2b :. M = —Ita 2b di lemf induced in larger coil I = M in smaller coil dt 2 b 2 current in the larger coil 2 = — Tta 2b INDUCED ELECTRIC FIELD x x Bin This electric field has following important properties : x (a) (b) (c) (d) (e) It is non conservative in nature. The line integral of E around a closed path is not zero. When a charge q goes once around the loop, the total work done on it by the electric field is equal to q times the emf. d(PB fE.dc e Hence dt Note, that the equation is valid only if the path around which we integrate is stationary. Because of symmetry, the electric field E at several points on the loop are shown in figure. Being a non-conservative field, the concept of potential has no meaning for such a field. This field is different from the electrostatic field produced by stationary charges (which is conservative in nature). The relation F =qE is still valid for this field.
  6. (f) This field can vary with time. A changing magnetic field acts as a source of electric field that cannot be produced with any static charge distribution. Example 6 : Solution : Find the magnitude of induced field " En " at a point r (>R) where a uniform but time varying dB magnetic field = b exists in a region of radius R. dt For the circular path of radius 'r' I fÉn.di I = En (kr) as I fÉn.di I = dt x x x x x x x x we get, 2 dB [No magnetic field for r > R] dt R2 dB 2r dt

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