{"id":39624,"date":"2023-01-29T10:00:41","date_gmt":"2023-01-29T04:30:41","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=39624"},"modified":"2023-01-30T09:51:06","modified_gmt":"2023-01-30T04:21:06","slug":"plus-two-physics-chapter-wise-previous-questions-chapter-4","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/plus-two-physics-chapter-wise-previous-questions-chapter-4\/","title":{"rendered":"Plus Two Physics Chapter Wise Previous Questions Chapter 4 Moving Charges and Magnetism"},"content":{"rendered":"

Plus Two Physics Chapter Wise Previous Questions Chapter 4 Moving Charges and Magnetism is part of Kerala Plus Two Physics Chapter Wise Previous Questions and Answers Kerala<\/a>. Here we have given Plus Two Physics Chapter Wise Questions and Answers Chapter 4 Moving Charges and Magnetism.<\/p>\n

Kerala Plus Two Physics Chapter Wise Previous Questions and Answers Chapter 4 Moving Charges and Magnetism<\/h2>\n

Question 1.
\nA particle of mass 1 x 10-26\u00a0<\/sup>kg and charge 1.6 x 10-19\u00a0<\/sup>C travelling with a velocity of 1.28 x 106\u00a0<\/sup>m\/s in the +X direction enters a region in which a uniform electric field \u00a7 and a uniform magnetic field of induction B are present such that Ex = Ey = 0 and Ez = – 102.4 kV\/m and Bx = Bz = 0, By = 8 x 10-2<\/sup> wb\/m2<\/sup>. The particle enters this region at the origin at time t = 0. (March – 2009)<\/span>
\na) Write the expression for the force experienced by the particle inside an electric field E
\nb) Find the force experienced by the particle in the electric field alone, in the above problem, (magnitude and direction)
\nc) Write the expression for the force experienced by the particle in a magnetic field.
\nd) Find the force experienced by the particle in the magnetic field alone, in the above problem, (magnitude and direction)
\ne) What is the resultant force on the particle in the above problem?
\nAnswer:
\n\"Plus
\ne) The two forces acting on charge (force due to electric field and another force due to magnetic field) are of the same magnitude but opposite in direction.<\/p>\n

Hence net force is zero.<\/p>\n

Question 1.
\nAmpere\u2019s circuital theorem is generally used to determine the magnetic field produced by a current-carrying element. (March – 2010)<\/span>
\na) State Ampere\u2019s circuital theorem.
\nb) Obtain an expression for the magnetic field produced by an infinitely long straight conductor using Ampere\u2019s circuital theorem.
\nc) A long straight conductor carries 35 ampere. Find the magnetic field produced due to this conductor at a point 20 cm away from the centre of the wire.
\nOR
\nA galvanometer is used to detect current in a circuit.
\na) State the working principle of a moving coil galvanometer.
\nb) How will you convert a galvanometer into (i) an ammeter and (ii) a voltmeter?
\nc) A galvanometer coil has a resistance of 12 ohms. It shows a full-scale deflection for a current of 3 mA. How will you convert this into a voltmeter of range 0-18V?
\nAnswer:
\na) The line integral of the magnetic field around a closed loop is PO times, the current enclosed by the loop.
\nb) Long straight conductor: Consider a long straight conductor carrying T ampere current. To find the magnetic field at \u2018P\u2019, we construct a circle of radius r (passing through P).
\n\"Plus
\nAccording to Ampere\u2019s circuital law we can write
\n\"Plus<\/p>\n

OR<\/p>\n

a) A current-carrying loop when placed in a magnetic field experiences a torque.
\nb) A galvanometer can be converted in to ammeter by connecting a small resistance in parallel to it. A galvanometer can be converted into a voltameter by connected a high resistance in series with it.<\/p>\n

c) lg<\/sub>\u00a0= 3 x 10-3\u00a0<\/sup>A, G = 12\u03a9, V= 18 V
\nThe resistance required to convert galvanometer in to voltmeter.
\n\"Plus
\nBy connecting a resistance R = 5988 in series with a galvanometer, we can convert a galvanometer into a voltmeter.<\/p>\n

Question 1.
\nA cyclotron is a device that accelerates charged particles for carrying out nuclear reactions. (Say – 2010)<\/span>
\na) With a suitable diagram, briefly explain its working principle and how charged particles get accelerated.
\nb) Find the energy of emergent protons (in MeV) coming from a cyclotron having dees of radius 2 metres, when a magnetic field of 0.8 T is applied. (Mass of proton = 1.67 x 10-27\u00a0<\/sup>kg)
\nAnswer:
\na) CYCLOTRON
\nUses
\nIt is a device used to accelerate particles to high energy.<\/p>\n

Principles
\nCyclotron is based on two facts<\/p>\n