{"id":34770,"date":"2022-11-19T16:47:51","date_gmt":"2022-11-19T11:17:51","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=34770"},"modified":"2022-11-23T16:18:31","modified_gmt":"2022-11-23T10:48:31","slug":"ts-inter-2nd-year-physics-notes-chapter-7","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-notes-chapter-7\/","title":{"rendered":"TS Inter 2nd Year Physics Notes Chapter 7 Moving Charges and Magnetism"},"content":{"rendered":"
Here students can locate TS Inter 2nd Year Physics Notes<\/a> 7th Lesson Moving Charges and Magnetism to prepare for their exam.<\/p>\n \u2192 Conclusion of Oersted states that moving charges produce a current and magnetic field in the surrounding space.<\/p>\n \u2192 If current carrying wire is perpendicular to the plane of the paper then magnetic field lines produced are concentric circles in the plane of the paper with the conductor at centre.<\/p>\n \u2192 If direction of current in a conductor is reversed then direction of magnetic field is also reversed.<\/p>\n \u2192 A static charge will produce only electric field whereas a moving charge will produce both magnetic field and electric field.<\/p>\n \u2192 Lorentz force: If a charge q is moving with a velocity V in electric (E\u0305) and magnetic fields (B) then total force on it is the sum of force due to electric field (Fele<\/sub>) and force due to magnetic field (Fmag<\/sub>) at that given point say V. \u2192 Force acting on a current carrying conductor placed in a magnetic field F = BIf sin \u03b8<\/p>\n \u2192 Motion of a charged particle in a magnetic field : if a charged particle q’ is moving perpendicularly in a magnetic field B\u0305 with a velocity ‘v’ then force due to magnetic field is always perpendicular to velocity V. So it describes a circular path. <\/p>\n \u2192 For a charged particle moving perpendicularly in a magnetic field mv2<\/sup>\/r = qvB. or radius of circular\/ helical path r = \\(\\frac{m v}{q B}\\)<\/p>\n \u2192 Velocity selector : If a charged particle is moving in a crossed electric (E\u0305) and magnetic fields (B\u0305) such that they will cancel each other then path of charged particle is undeviated, i.e., when qE = qvB \u2192 Cyclotron : Cyclotron is a charged particle accelerator. \u2192 From Biot – Savart’s law magnetic field due to a current carrying conductor at a given point P’ is given by dB = \\(\\frac{\\mu_0}{4 \\pi} \\frac{I d l \\sin \\theta}{\\mathbf{r}^2}\\)<\/p>\n \u2192 From Ampere’s circuital law the total mag\u00acnetic field coming out of a current carrying conductor is p0 times greater than the cur\u00acrent flowing through it. \u2192 Solenoid : A solenoid consists of a long wire wound on an insulating hollow cylinder in the form of helix. \u2192 Toroid : Toroid is a coiled coil. \u2192 Torque on a current loop \u03c4 = IAB sin \u03b8 \u2192 Magnetic moment of current loop M\u0305 = IA\u0305; OR nIA\u0305 \u2192 For a revolving electron : Current I = e\/T \u2192 Moving Coil Galvanometer (M.C.G) : <\/p>\n \u2192 To convert galvanometer into ammeter shunt to be added Rs<\/sub> = \\(\\frac{G}{n-1}\\). \u2192 To convert galvanome fer into voltmeter series resistance to be added Rs<\/sub> = \\(\\frac{\\mathrm{V}}{\\mathrm{I}_{\\mathrm{g}}}\\) – RG<\/sub> Here students can locate TS Inter 2nd Year Physics Notes 7th Lesson Moving Charges and Magnetism to prepare for their exam. TS Inter 2nd Year Physics Notes 7th Lesson Moving Charges and Magnetism \u2192 Conclusion of Oersted states that moving charges produce a current and magnetic field in the surrounding space. \u2192 If current carrying … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[1],"tags":[],"yoast_head":"\nTS Inter 2nd Year Physics Notes 7th Lesson Moving Charges and Magnetism<\/h2>\n
\nF = q [ E(r) + v\u0305 \u00d7 B\u0305 (r)] = Fele<\/sub> + Fmag<\/sub>
\nThis is known as Lorentz force.<\/p>\n
\nSince force and displacement are perpendicular no work is done.<\/p>\n
\n\u21d2 v = \\(\\frac{E}{B}\\)
\nThe ratio of \\(\\frac{E}{B}\\) for undeviation condition B is called velocity selector.<\/p>\n
\nCyclotron frequency \u03c5c<\/sub> = \\(\\frac{\\mathrm{qB}}{2 \\pi \\mathrm{m}}\\) is independent of velocity of charged particle.<\/p>\n
\n\u222eB. dI = \u00b50<\/sub>I<\/p>\n
\nNet magnetic field is the vector sum of fields due to all turns.
\nMagnetic field along the axis of solenoid B= n0nI<\/p>\n
\nIt consists of a large number of turns of wire which are closely wound on a ring. Or Toroid is a solenoid bent into the form of a ring.
\nIn a toroid magnetic field B = \\(\\frac{\\mu_0 \\mathrm{NI}}{2 \\pi \\mathrm{r}}\\)
\nWhere N = 2\u03c0rn = perimeter of toroid \u00d7 number of turns per unit length.
\nCurrent loop:<\/p>\n
\nFor a loop of n turns torque \u03c4 = n IAB sin \u03b8
\nHere sin \u03b8 is the angle between Area vector A\u0305 and direction of magnetic field B\u0305.<\/p>\n
\nTorque \u03c4 = M\u0305 x B\u0305<\/p>\n
\nMagnetic moment \u03bc1<\/sub> = \\(\\frac{\\mathrm{e}}{2 \\mathrm{~m}_{\\mathrm{e}}}\\)(me<\/sub>vr) = \\(\\frac{\\mathrm{e}}{2 \\mathrm{~m}_{\\mathrm{e}}}\\)I ;
\nTime period T = \\(\\frac{2 \\pi r}{v}\\)
\nGyromagnetic ratio \\(\\frac{\\mathrm{e}}{2 \\mathrm{~m}_{\\mathrm{e}}}\\)
\nAngular momentum I = \\(\\frac{\\mathrm{nh}}{2 \\pi}\\)<\/p>\n
\nTorque on the coil \u03c4 = nlAB deflection \u03a6 = \\(\\left(\\frac{\\mathrm{NAB}}{\\mathbf{k}}\\right)\\)I, k = torsional constant of spring.<\/p>\n
\nWhere n = \\(\\left(\\frac{i}{i_g}\\right)=\\frac{\\text { new current }}{\\text { old current }}\\)
\n= \\(\\frac{\\text { current to be measured }}{\\text { current permitted through galvanometer }}\\)<\/p>\n
\nWhere Ig<\/sub> = Current permitted through galvanometer
\nV = Voltage to be measured.
\nRG<\/sub> = Resistance of galvanometer.<\/p>\n","protected":false},"excerpt":{"rendered":"