{"id":36066,"date":"2022-11-26T12:44:49","date_gmt":"2022-11-26T07:14:49","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=36066"},"modified":"2022-11-26T12:44:49","modified_gmt":"2022-11-26T07:14:49","slug":"ts-inter-2nd-year-physics-study-material-chapter-9","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/","title":{"rendered":"TS Inter 2nd Year Physics Study Material Chapter 9 Electromagnetic Induction"},"content":{"rendered":"

Telangana TSBIE\u00a0TS Inter 2nd Year Physics Study Material<\/a> 9th Lesson Electromagnetic Induction Textbook Questions and Answers.<\/p>\n

TS Inter 2nd Year Physics Study Material 9th Lesson Electromagnetic Induction<\/h2>\n

Very Short Answer Type Questions<\/span><\/p>\n

Question 1.
\nWhat did the experiments of Faraday and Henry show?
\nAnswer:<\/p>\n

    \n
  1. Faraday and Henry experiments showed that the relative motion between the magnet and a coil is responsible for generation of electric current in the coil.<\/li>\n
  2. The relative motion is not an absolute requirement to induce the current in a coil. If the current in a coil changes then also emf is induced in the nearby coil.<\/li>\n<\/ol>\n

    Question 2.
    \nDefine magnetic flux.
    \nAnswer:
    \nMagnetic flux :
    \nThe number of magnetic field lines crossing unit area when placed perpendicular to the field is defined as “Magnetic flux”.
    \nMagnetic flux, \u03a6 = \\(\\overline{\\mathrm{B}}.\\overline{\\mathrm{A}}\\) = B A cos \u03b8<\/p>\n

    Question 3.
    \nState Faraday’s law of electromagnetic induction.
    \nAnswer:
    \nFaraday’s law of Induction :
    \nThe rate of change of magnetic flux through a circular coil induces emf in it.
    \n\"TS<\/p>\n

    Question 4.
    \nState Lenz’s law. [TS Mar. 19, June 15]
    \nAnswer:
    \nLenz’s Law :
    \nThe polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produces current in that coil.<\/p>\n

    Question 5.
    \nWhat happens to the mechanical energy (of motion) when a conductor is moved in a uniform magnetic field?
    \nAnswer:
    \nWhen a condutor is moved in a magnetic field
    \nPower of this motion, P = Fv F = Bil
    \n\"TS
    \nThe work done is in the form of mechanical energy.
    \nThis is dissipated into the form of joule heat.
    \n\u2234 Joule heat = Power Pj<\/sub> = I\u00b2r = \\(\\frac{B^2l^2v^2}{r}\\)<\/p>\n

    \"TS<\/p>\n

    Question 6.
    \nWhat are Eddy currents? [TS Mar. ’19; AP June ’15]
    \nAnswer:
    \nEddy currents :
    \nWhen large pieces of conductors are subjected to changing magnetic flux then current is induced in them. These induced currents are called \u201dEddy currents”.<\/p>\n

    Eddy currents will oppose the motion of the coil (or) they oppose the change in magnetic flux.<\/p>\n

    Question 7.
    \nDefine ‘inductance’.
    \nAnswer:
    \nInductance :
    \nThe process of producing emf in a coil due to changing current in that coil or in a coil nearby it is called “Inductance”.<\/p>\n

    Flux associated with a coil \u03a6B<\/sub> is proportional to current i.e., \u03a6B<\/sub> \u221d I
    \n\"TS
    \nThis constant of proportionality is called Inductance.<\/p>\n

    Question 8.
    \nWhat do you understand by ‘self induetance’? [AP & TS June ’15]
    \nAnswer:
    \nSelf inductance :
    \nIf emf is induced in a single isolated coil due to change of flux in that coil by means of changing current through that coil then that phenomenon is called “Self Inductance L”.
    \nIn Self inductance, \u03b5 = -L\\(\\frac{dI}{dt}\\)<\/p>\n

    Short Answer Questions<\/span><\/p>\n

    Question 1.
    \nObtain an expression for the emf induced across a conductor which is moved in a uniform magnetic field which is perpendicular to the plane of motion. [TS May ’16]
    \nAnswer:
    \nLet a conductor of length T is moving with a velocity “v” in a uniform and time independent magnetic field B.<\/p>\n

    Consider a rectangular metallic frame PQRS in which the side PQ is free to move without friction.
    \n\"TS
    \nLet PQ move with a velocity ‘v’ in a perpendicular magnetic field B.<\/p>\n

    Magnetic flux in the loop \u03a6B<\/sub> = Bl . x, where x = RQ a time changing quantity.
    \n\"TS
    \nHere \\(\\frac{dx}{dt}\\) = -v = Velocity of the rod.
    \nInduced emf, \u03b5 = Blv is called Motional emf.<\/p>\n

    Question 2.
    \nDescribe the ways in which Eddy currents are used to advantage. [AP Mar. ’19, ’18, ’17, ’16, ’15, May ’18, ’17, ’16; TS Mar. ’18, ’15. May \u201917]
    \nAnswer:
    \nAdvantages of Eddy currents:<\/p>\n

      \n
    1. Electromagnetic breaking : In some electrically powered trains strong electromagnets are placed above rails. When these electromagnets are activated eddy currents induced in rails will oppose motion of train. These breaks are smooth.<\/li>\n
    2. In galvanometers, a fixed core is made with non-magnetic material. When coil oscillates eddy currents induced in core will oppose the motion. As a result, the coil will come to rest quickly.<\/li>\n
    3. In induction furnaces high frequency oscillating currents are passed through a coil which surrounds the metal to be melted. These currents will produce eddy currents in the metal and it is heated sufficiently to melt it.<\/li>\n
    4. In electric power meters a metal disc is made to rotate due to eddy currents with some arrangement. Rotation of this disc is made to measure power consumed.<\/li>\n<\/ol>\n

      \"TS<\/p>\n

      Question 3.
      \nObtain an expression for the mutual inductance of two long co-axial solenoids.
      \nAnswer:
      \nConsider two long solenoids S1<\/sub> and S2<\/sub> each of length ‘l’, radius r1<\/sub> and r2<\/sub> and number of turns n1<\/sub> and n2<\/sub> respectively. When a current I2<\/sub> is sent through S2<\/sub> it will set up a magnetic flux \u03a61<\/sub> through S1<\/sub>.
      \n\"TS<\/p>\n

      Flux linkage with S1<\/sub> is N1<\/sub> \u03a61<\/sub> = M12<\/sub>I2<\/sub> where
      \nM12<\/sub> is mutual inductance between the coils.
      \nBut \u03a61<\/sub> = N1<\/sub>A1<\/sub>B where N1<\/sub> = n1<\/sub>l ; A = \u03c0r\u00b21<\/sub> and B = \u00b50<\/sub>n2<\/sub>I2<\/sub>.
      \n\u2234 N1<\/sub>\u03a61<\/sub> =(n1<\/sub>l)(\u03c0r\u00b21<\/sub> )(\u00b50<\/sub>n2<\/sub>I2<\/sub>) = \u00b50<\/sub>n1<\/sub>n2<\/sub>r\u00b21<\/sub> …………. (1)<\/p>\n

      This approximation is highly valid when l > > r2<\/sub>.<\/p>\n

      If current I1<\/sub> is passed through S1<\/sub> then
      \nN2<\/sub>\u03a62<\/sub> = M21<\/sub>I1<\/sub> where \u03a62<\/sub> = N2<\/sub>A2<\/sub>B and
      \nB = \u00b50<\/sub>n1<\/sub>I1<\/sub> and N2<\/sub> = n2<\/sub>l.
      \n\u2234 N2<\/sub>\u03a62<\/sub> =(n2<\/sub>l)(\u03c0r\u00b21<\/sub> )(\u00b50<\/sub>n1<\/sub>I1<\/sub>) = \u00b50<\/sub>n1<\/sub>n2<\/sub>\u03c0r\u00b21<\/sub>l …………. (2)<\/p>\n

      From eq. (1) & (2) Mutual inductance bet-ween co-axial solenoids M12<\/sub> = M21<\/sub>. If the solenoid is on a core of permeability \u00b5r<\/sub> then
      \n[M12<\/sub> = M21<\/sub> = \u00b50<\/sub>\u00b5r<\/sub>n1<\/sub>n2<\/sub>\u03c0r\u00b21<\/sub>l]
      \nMutual inductance of a pair of coils or solenoids etc., depends on seperation between them and also on their orientaton.<\/p>\n

      Question 4.
      \nObtain an expression for the magnetic energy stored in a solenoid in terms of the magnetic field, area and length of the solenoid.
      \nAnswer:
      \nWhen current is passed through a single isolated coil or solenoid changing magnetic flux can be developed by changing current through it. This changing flux will induce emf in that coil.
      \nThis phenomenon is called self induction (L).
      \nFlux linkage N\u03a6B<\/sub> \u221d I or N\u03a6B<\/sub> = L . I
      \nInduced emf, \u03b5 = \\(\\frac{d}{dt}\\)(N\u03a6B<\/sub>) = -L\\(\\frac{dI}{dt}\\) ………. (1)<\/p>\n

      – ve sign indicates that induced emf will always oppose the flux changes in that coil (or) solenoid.<\/p>\n

      Let length of solenoid is ‘l’, area of cross section = A
      \nthen N\u03a6B<\/sub> = (nl)(\u00b50<\/sub>nI) (I) (\u2235 \u03a6B<\/sub> =n\u00b50<\/sub>I) ……….. (2)
      \nand total number of turns N = n \u00d7 l.
      \ni.e., turns per unit length ‘n’ \u00d7 length of solenoid ‘l’.
      \n\u2234 L = \\(\\frac{\\mathrm{N} \\phi_{\\mathrm{B}}}{\\mathrm{I}}\\) = \u00b50<\/sub>n\u00b2Al …………. (3)<\/p>\n

      This self induced emf also called back emf will oppose any change in current in the coil. So to drive current in the circuit we must do some work.
      \nRate of work done = \\(\\frac{dW}{dt}\\) = |\u03b5|I = LI.\\(\\frac{dI}{dt}\\) …………. (4)
      \n\u2234 Energy required to send the currrent or Energy stored in inductor
      \n\"TS<\/p>\n

      Long Answer Questions<\/span><\/p>\n

      Question 1.
      \nOutline the path-breaking experiments of Faraday and Henry and highlight the contributions of these experiments to our understanding of electromagnetism.
      \nAnswer:
      \nFaraday and Henry conducted a series of experiments to understand electromagnetic inductioin.
      \n\"TS<\/p>\n

      First Experiment:
      \nIn this experiment, a galvanometer is connected to a coil. A magnet is moved towards the coil. They observed that current is flowing in the coil when magnet is in motion.<\/p>\n

      The direction of induced current is in opposite direction when the direction of motion of magnet is changed.<\/p>\n

      So they concluded that relative motion between magnet and coil is responsible for generation of electric current in the coil.<\/p>\n

      Second Experiment:
      \nIn this experiment a steady current is passed through one coil with the help of a battery and the second coil is connected to a galvanometer. When one of the coil is moved then current is induced in the coil. This current losts as far as there is relative motion between them.<\/p>\n

      Again they concluded that relative motion between the coils is responsible for induced electric current.<\/p>\n

      Third Experiment:
      \nIn this experiment, they connected one coil to a battery and a tapping key to make and break electric contact in that coil. The second coil is placed near the first coil. When electric contact is established current is induced in the second coil and momentary deflection is observed in galvanometer.<\/p>\n

      When electric contact is breaked again they got deflection in galvanometer in opposite direction.<\/p>\n

      So they concluded that it is not the relative motion between the coil and magnet or relative motion between the coils that induces the current. The changing magnetic flux is responsible for induced emf or current in the coil.<\/p>\n

      Finally Faraday proposed that induced emf \u03b5 = \\(\\frac{\\mathrm{d} \\phi_{\\mathrm{B}}}{\\mathrm{dt}}\\) \u21d2 \u03b5 = N.\\(\\frac{\\mathrm{d} \\phi_{\\mathrm{B}}}{\\mathrm{dt}}\\)
      \nBut from Lenz’s explanation he corrected this equation as \u03b5 = –\\(\\frac{\\mathrm{d} \\phi_{\\mathrm{B}}}{\\mathrm{dt}}\\) \u21d2 \u03b5 = -N\\(\\frac{\\mathrm{d} \\phi_{\\mathrm{B}}}{\\mathrm{dt}}\\)<\/p>\n

      \"TS<\/p>\n

      Question 2.
      \nDescribe the working of a AC generator with the aid of a simple diagram and necessary expressions.
      \nAnswer:
      \nAC generator consists of a coil of N turns placed in a magnetic field B produced by magnetic poles when the coil is rotated its effective area changes so flux linked with the coil changes. This changing flux will induce emf in the coil.<\/p>\n

      Electric generator converts mechanical energy into electrical energy.
      \n\"TS
      \nFlux associated with the coil,
      \n\u03a6B<\/sub> = (\\(\\overline{\\mathrm{B}}\\) \u2022 \\(\\overline{\\mathrm{A}}\\)) = BA cos \u03b8, where \u03b8 = \u03c9t
      \nInduced emf, \u03b5 = -N\\(\\frac{\\mathrm{d} \\phi_{\\mathrm{B}}}{\\mathrm{dt}}\\) = -NBA\\(\\frac{d}{dt}\\)cos \u03c9t
      \n\u2234 Induced emf, \u03b5 = -NBA\u03c9 sin \u03c9t
      \nThe term NBA\u03c9 is called maximum emf produced (\u03b5m<\/sub>).
      \n\u2234 \u03b5m<\/sub> = NBA\u03c9
      \n\u03b5 = \u03b5m<\/sub> sin \u03c9t
      \nInduced emf at any time E = \u03b5m<\/sub> sin \u03c9t
      \nWhen \u03b8 = 0, Induced emf is zero i.e., \u03b5 = 0.
      \nThe induced emf is maximum when \u03b8 = 90\u00b0
      \ni.e., the plane of the coil is perpendicular
      \nto magnetic field.
      \nWhen \u03b8 = 90\u00b0 \u21d2 emf \u03b5 = \u03b5m<\/sub>
      \nWhen \u03b8 = 180\u00b0 \u21d2 induced emf \u03b5 = 0.
      \nWhen \u03b8 = 270\u00b0 \u21d2 induced emf \u03b5 = – \u03b5m<\/sub>.
      \nagain for \u03b8 = 360\u00b0 \u21d2 emf \u03b5 = 0.
      \n\u2234 The induced emf varies sinusoidally in AC generator.
      \n\"TS<\/p>\n

      The coil is mounted on a rotor shaft. The axis of rotation of coil is perpendicular to magnetic field. The coil is connected to external circuit by means of slip rings and brushes.<\/p>\n

      Induced emf at any time is given by \u03b5 = \u03b5m<\/sub> sin \u03c9t = \u03b5m<\/sub> sin 2\u03c0\u03c5t.<\/p>\n

      Depending on the method of supplying mechanical energy to rotate shaft these AC generators are classified as 1) Hydroelectric generators, 2) Thermal generators and 3) Nuclear generators.<\/p>\n

      Exercises<\/span><\/p>\n

      Question 1.
      \nObtain an expression for the emf induced across a conductor which is moved in a uniform magnetic field which is perpendicular to the plane of motion. [AP May ’14]
      \nAnswer:
      \nLet a conductor of length ‘l’ is moving with a velocity “v” in a uniform and time independent magnetic field B.<\/p>\n

      Consider a rectangular metallic frame PQRS in which the side PQ is free to move without friction.<\/p>\n

      Let the wire PQ is moved with a velocity ‘v’ in a perpendicular magnetic field B.
      \n\"TS
      \nMagnetic flux in the loop \u03a6B<\/sub> = Bl . x, where x = RQ a time changing quantity.
      \n\"TS
      \nHere \\(\\frac{dx}{dt}\\) = -v = Velocity of the rod.
      \nInduced emf, \u03b5 = Blv is called Motional emf.<\/p>\n

      Question 2.
      \nCurrent in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self inductance of the circuit. [AP Mar. 14; TS Mar. 16]
      \nAnswer:
      \nInitial current, I1<\/sub> = 5.0 A ;
      \nFinal current, I2<\/sub> = 0.0 A ;
      \nChange in current, dl = I1<\/sub> – I2<\/sub> = 5 A
      \nTime taken for the change, t = 0.1 s;
      \nAverage emf, \u03b5 = 200 V
      \nFor self-inductance (L) of the coil, we have the relation for average emf as
      \n\"TS
      \nHence, the self induction of the coil is 4H.<\/p>\n

      Question 3.
      \nA pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? [TS May ’18, Mar. ’17]
      \nAnswer:
      \nMutual inductance of a pair of coils, \u00b5 = 1.5 H;
      \nInitial current, I1<\/sub> = 0 A
      \nFinal current I2<\/sub> = 20 A ;
      \nChange in current, dl – I2<\/sub> – I1<\/sub> = 20 – 0 = 20 A
      \nTime taken for the change, t = 0.5 s
      \n\"TS
      \nWhere d\u03a6 is the change in the flux linkage with the coil.
      \nEquating equations (1) and (2), we get
      \n\\(\\frac{\\mathrm{d} \\phi}{\\mathrm{dt}}\\) = \u00b5\\(\\frac{dI}{dt}\\) ; d\u03a6 = 1.5 \u00d7 (20) = 30 Wb
      \nHence, the change in the flux linkage is 30 Wb.<\/p>\n

      \"TS<\/p>\n

      Question 4.
      \nA jet plane is travelling towards west at a speed of 1800 km\/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 \u00d7 10-4<\/sup> T and the dip angle is 30\u00b0.
      \nAnswer:
      \nSpeed of the jet plane, v = 1800 km\/h = 500 m\/s ;
      \nWing span of jet plane, l = 25 m
      \nEarth’s magnetic field strength,
      \nB = 5.0 \u00d7 10-4<\/sup> T ; Angle of dip, \u03b4 = 30\u00b0<\/p>\n

      Vertical component of Earth’s magnetic field,
      \nBv<\/sub> = B sin \u03b4 = 5 \u00d7 10-4<\/sup> sin 30\u00b0 = 2.5 \u00d7 10-4<\/sup>T<\/p>\n

      Voltage difference between the ends of the wing can be calculated as
      \n\u03b5 = (Bv<\/sub>) \u00d7 l \u00d7 v = 2.5 \u00d7 10-4<\/sup> \u00d7 25 \u00d7 500 = 3.125 V<\/p>\n

      Hence, the voltage difference developed between the ends of the wings is 3.125 V.<\/p>\n","protected":false},"excerpt":{"rendered":"

      Telangana TSBIE\u00a0TS Inter 2nd Year Physics Study Material 9th Lesson Electromagnetic Induction Textbook Questions and Answers. TS Inter 2nd Year Physics Study Material 9th Lesson Electromagnetic Induction Very Short Answer Type Questions Question 1. What did the experiments of Faraday and Henry show? Answer: Faraday and Henry experiments showed that the relative motion between the … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[26],"tags":[],"yoast_head":"\nTS Inter 2nd Year Physics Study Material Chapter 9 Electromagnetic Induction - TS Board Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"TS Inter 2nd Year Physics Study Material Chapter 9 Electromagnetic Induction - TS Board Solutions\" \/>\n<meta property=\"og:description\" content=\"Telangana TSBIE\u00a0TS Inter 2nd Year Physics Study Material 9th Lesson Electromagnetic Induction Textbook Questions and Answers. TS Inter 2nd Year Physics Study Material 9th Lesson Electromagnetic Induction Very Short Answer Type Questions Question 1. What did the experiments of Faraday and Henry show? Answer: Faraday and Henry experiments showed that the relative motion between the ... Read more\" \/>\n<meta property=\"og:url\" content=\"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/\" \/>\n<meta property=\"og:site_name\" content=\"TS Board Solutions\" \/>\n<meta property=\"article:published_time\" content=\"2022-11-26T07:14:49+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/11\/TS-Inter-2nd-Year-Physics-Study-Material-Chapter-9-Electromagnetic-Induction-1.png\" \/>\n<meta name=\"author\" content=\"Srinivas\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Srinivas\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"11 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/\"},\"author\":{\"name\":\"Srinivas\",\"@id\":\"https:\/\/tsboardsolutions.com\/#\/schema\/person\/1d1aa14a8abcb28fa5e85127be78dcea\"},\"headline\":\"TS Inter 2nd Year Physics Study Material Chapter 9 Electromagnetic Induction\",\"datePublished\":\"2022-11-26T07:14:49+00:00\",\"dateModified\":\"2022-11-26T07:14:49+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/\"},\"wordCount\":2159,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/tsboardsolutions.com\/#organization\"},\"articleSection\":[\"TS Inter 2nd Year\"],\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/\",\"url\":\"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/\",\"name\":\"TS Inter 2nd Year Physics Study Material Chapter 9 Electromagnetic Induction - TS Board Solutions\",\"isPartOf\":{\"@id\":\"https:\/\/tsboardsolutions.com\/#website\"},\"datePublished\":\"2022-11-26T07:14:49+00:00\",\"dateModified\":\"2022-11-26T07:14:49+00:00\",\"breadcrumb\":{\"@id\":\"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/tsboardsolutions.com\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"TS Inter 2nd Year Physics Study Material Chapter 9 Electromagnetic Induction\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/tsboardsolutions.com\/#website\",\"url\":\"https:\/\/tsboardsolutions.com\/\",\"name\":\"TS Board Solutions\",\"description\":\"Telangana TS Board Textbook Solutions for Class 6th, 7th, 8th, 9th, 10th, Inter 1st & 2nd Year\",\"publisher\":{\"@id\":\"https:\/\/tsboardsolutions.com\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/tsboardsolutions.com\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"Organization\",\"@id\":\"https:\/\/tsboardsolutions.com\/#organization\",\"name\":\"TS Board Solutions\",\"url\":\"https:\/\/tsboardsolutions.com\/\",\"logo\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/tsboardsolutions.com\/#\/schema\/logo\/image\/\",\"url\":\"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/10\/cropped-TS-Board-Solutions-1.png\",\"contentUrl\":\"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/10\/cropped-TS-Board-Solutions-1.png\",\"width\":488,\"height\":40,\"caption\":\"TS Board Solutions\"},\"image\":{\"@id\":\"https:\/\/tsboardsolutions.com\/#\/schema\/logo\/image\/\"}},{\"@type\":\"Person\",\"@id\":\"https:\/\/tsboardsolutions.com\/#\/schema\/person\/1d1aa14a8abcb28fa5e85127be78dcea\",\"name\":\"Srinivas\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/tsboardsolutions.com\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/10a27cfafdf21564c686b80411336ece?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/10a27cfafdf21564c686b80411336ece?s=96&d=mm&r=g\",\"caption\":\"Srinivas\"},\"url\":\"https:\/\/tsboardsolutions.com\/author\/srinivas\/\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"TS Inter 2nd Year Physics Study Material Chapter 9 Electromagnetic Induction - TS Board Solutions","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/","og_locale":"en_US","og_type":"article","og_title":"TS Inter 2nd Year Physics Study Material Chapter 9 Electromagnetic Induction - TS Board Solutions","og_description":"Telangana TSBIE\u00a0TS Inter 2nd Year Physics Study Material 9th Lesson Electromagnetic Induction Textbook Questions and Answers. TS Inter 2nd Year Physics Study Material 9th Lesson Electromagnetic Induction Very Short Answer Type Questions Question 1. What did the experiments of Faraday and Henry show? Answer: Faraday and Henry experiments showed that the relative motion between the ... Read more","og_url":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/","og_site_name":"TS Board Solutions","article_published_time":"2022-11-26T07:14:49+00:00","og_image":[{"url":"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/11\/TS-Inter-2nd-Year-Physics-Study-Material-Chapter-9-Electromagnetic-Induction-1.png"}],"author":"Srinivas","twitter_card":"summary_large_image","twitter_misc":{"Written by":"Srinivas","Est. reading time":"11 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/#article","isPartOf":{"@id":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/"},"author":{"name":"Srinivas","@id":"https:\/\/tsboardsolutions.com\/#\/schema\/person\/1d1aa14a8abcb28fa5e85127be78dcea"},"headline":"TS Inter 2nd Year Physics Study Material Chapter 9 Electromagnetic Induction","datePublished":"2022-11-26T07:14:49+00:00","dateModified":"2022-11-26T07:14:49+00:00","mainEntityOfPage":{"@id":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/"},"wordCount":2159,"commentCount":0,"publisher":{"@id":"https:\/\/tsboardsolutions.com\/#organization"},"articleSection":["TS Inter 2nd Year"],"inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/","url":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/","name":"TS Inter 2nd Year Physics Study Material Chapter 9 Electromagnetic Induction - TS Board Solutions","isPartOf":{"@id":"https:\/\/tsboardsolutions.com\/#website"},"datePublished":"2022-11-26T07:14:49+00:00","dateModified":"2022-11-26T07:14:49+00:00","breadcrumb":{"@id":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-9\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/tsboardsolutions.com\/"},{"@type":"ListItem","position":2,"name":"TS Inter 2nd Year Physics Study Material Chapter 9 Electromagnetic Induction"}]},{"@type":"WebSite","@id":"https:\/\/tsboardsolutions.com\/#website","url":"https:\/\/tsboardsolutions.com\/","name":"TS Board Solutions","description":"Telangana TS Board Textbook Solutions for Class 6th, 7th, 8th, 9th, 10th, Inter 1st & 2nd Year","publisher":{"@id":"https:\/\/tsboardsolutions.com\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/tsboardsolutions.com\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Organization","@id":"https:\/\/tsboardsolutions.com\/#organization","name":"TS Board Solutions","url":"https:\/\/tsboardsolutions.com\/","logo":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/tsboardsolutions.com\/#\/schema\/logo\/image\/","url":"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/10\/cropped-TS-Board-Solutions-1.png","contentUrl":"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/10\/cropped-TS-Board-Solutions-1.png","width":488,"height":40,"caption":"TS Board Solutions"},"image":{"@id":"https:\/\/tsboardsolutions.com\/#\/schema\/logo\/image\/"}},{"@type":"Person","@id":"https:\/\/tsboardsolutions.com\/#\/schema\/person\/1d1aa14a8abcb28fa5e85127be78dcea","name":"Srinivas","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/tsboardsolutions.com\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/10a27cfafdf21564c686b80411336ece?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/10a27cfafdf21564c686b80411336ece?s=96&d=mm&r=g","caption":"Srinivas"},"url":"https:\/\/tsboardsolutions.com\/author\/srinivas\/"}]}},"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/posts\/36066"}],"collection":[{"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/comments?post=36066"}],"version-history":[{"count":1,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/posts\/36066\/revisions"}],"predecessor-version":[{"id":36128,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/posts\/36066\/revisions\/36128"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/media?parent=36066"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/categories?post=36066"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/tags?post=36066"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}