Ans: 1) A is paramagnetic material. 2) B is diamagnetic material. THREE MARKS QUESTIONS 1. State law for magnetic field produced at a point due to a small current element. How will you find the direction of magnetic field? Ans. Consider a current element dl of a conductor carrying current I. let P be a point having . Let and position vector r with respect to current element dl be the angle between dl r According to Biot-Savart law, the magnetic field dB at point P due to current element is (i) dB I (ii) dB dl (iii) dB sin (iv) dB 1 r 2 Combining all the four factors, we get dB I dl r 2 dB = K I dl r 2 The proportionality constant K depends on the medium between point P and the current element and the system of units chosen. For free space and in SI units, K 0 10 T dB I dl r 2 and , so from the above As the direction of dB is perpendicular to the plane of dl r sin sin 4 7
If s be the position vector of point P relative to the elementdl, then from Biot-Savart law, the field at point P due to the current element is dB I dl r 2 = , therefore Since dl s, i.e., dB Idl s2 The field dB lies in the plane of paper and is perpendicular to s , as shown by PQ. be the angle between OP and CP. Then dB can be resolved into two rectangular components. 1. dB the axis, 2. perpendicular to the axis. For any two diametrically opposite elements of the loop, the components perpendicular to the axis of the loop will be equal and opposite and will cancel out. Their axial components will be in the same direction, i.e., along CP and get added up. Therefore, total magnetic field at the point P in the direction CP is = sin a and dB Idl s s2 B .Idl.a Since and I are constant, and s and aare same for all points on the circular loop, we have = 0Ia I [ 3 3 2 s 3 As, s 2 B 0 I a2 2 2 (r 4 4 4
B.dl B.dl B dl cos B B l -------(2) a a a Where, l = length of the side ab of the rectangular loop abcd. Now calculation for total current threading the loop abcd Let number of turns per unit length of the solenoid = n Then number of turns in length l of the solenoid = nl Total current threading the loop abcd = nlI----------(3) Putting the value of (2) and(3) in (1) We get Bl = or B Q 4. Discuss the motion of a charged particle in a uniform magnetic field with initial velocity parallel to the field, (ii) perpendicular to the magnetic field and (iii) at an arbitrary angle with the field direction. Ans. When a charged particle having charge q and velocity v enters a magnetic field B , it experiences a force F q(v The direction of this force is perpendicular to both v and B . The magnitude of this force is F = qv B sinθ Following three cases are possible: 1. When the initial velocity is parallel to the magnetic field: Here =0 , So F = qvB sin0 = 0. Thus the parallel magnetic field does not exert any force on the moving charged particle. The charged particle will continue to move along the line of force. nI
2. When the initial velocity is perpendicular to the magnetic field: Figure shows a magnetic field B directed normally into the plane of paper, as shown by small crosses. A charge +qis projected with a speed v in the plane of the paper. The velocity is perpendicular to the magnetic field. A force F = qvB acts on the particle perpendicular to both v and B . This force continuously deflects the particle sideways without changing its speed and the particle will move along a circle perpendicular to the field. Thus the magnetic force provides the centripetal force. Let r be the radius of the circular path. Centripetal force, mv2 = Magnetic force, qvB r r = mv qB Period of revolution, T= . mv v v qB qB The frequency of revolution is f c 1 qB T This frequency is independent of v and r and is called cyclotron frequency. 3. When the initial velocity makes an arbitrary angle with the field direction: Consider a charged particle q entering a uniform magnetic field B with velocity inclined at an angle with the direction of B ,as shown in figure. m m
Displacement current is that current which appears in a region in which the electric field (and hence electric flux) is changing with time. Note- We have ID = ε0 dΦ/dt = ε0d(EA) /dt = ε0d[(q/ ε0A)A] /dt = dq /dt = I 2. Modified Ampere’s circuital Law ∮B.dl = µ0 (I + ε0 dΦ/dt ) 3. Learn only one order either increasing wavelength or frequency .other calculated by c= ν λ ,velocity of light c=3x108m/s S.No. Name Frequency Range (Hz) Wavelength Range Production Uses 1. Gamma rays 1019 – 1023 10-11 to Emitted by radioactive nuclei, Produced in nuclear reaction In medicine, to destroy cancer cells. 2. X – rays 1016 - 1020 10-8 to Generated by bombarding a metal target by high energy electron Used as diagnostic tool in medicine, to study crystal structures 3. Ultraviolet rays 1015 – 1017 (4 × 10-7 to 6 × 10-10) m Produced by special lamps & very hot bodies (sun). For eye surgery, to kill germs in water purifiers. 4. Visible rays × 1014 – × 1014 700 – 400 nm Jumping of electrons in higher orbits Provide us information about the world. 5. Infrared rays (heat waves) 1012 - 1014 1mm - 700 nm Produced by hot bodies and molecules. Infrared detectors used in earth satellite, used in green house to keep plants warm. 6. microwaves 1010 - 1012 0.1 – 1 mm Produced by special vacuum tubes (klystrons, gun diode & magnetrons) Microwave oven, for radar system in aircraft navigation. 7. Radio waves 10 - 109 > 0.1 m Produced by accelerated motion of charges in conducting wires. In radio & television communication system, in 10-
IMPORTANT FORMULAE Mirror Formula Magnification (m) = I/O = -v/u Rrefractive index Apparent Shift Relation between critical angle and refractive index . Refraction at spherical surface Lens formula Lens makers formula Magnification (m) = I/O = v/u Power of lens Combination of lens Relation between i, e and Prism formula Magnifying power of simple microscope image at d) m = d/f Magnifying power of simple microscope (final image at d) (final image at infinity) Magnifying power of Telescope Length of tube Condition for constructive interference (
basically a p-n junction operating in a photovoltaic mode without external bias. Working: When light photons fall at the junction electron-hole pairs are generated. those more in opposite direction due to junction field. These charges accumulate at the two sides of the junction and photo voltage is developed. Use: It is used in calculators etc.SOME IMPORTANT QUESTIONS FROM PREVIOUS YEAR PAPERS (CHAPTER-ELECTRONIC DEVICES) 1.In a semiconductor the concentration of electorns is 8x1013cm-3 and that of holes is 5x1012cm-3. Is a p-type or n-type semiconductor? Ans : As concentration of electrons is more than holes, the given extrinsic semi conductor is n-type. 2.The energy gaps in the energy band diagrams of a conductor, semiconductor and insulator are E1, E2 and E3. Arrange them in increasing order. Ans: the energy gap in a conductor is zero, in a semiconductor is ≈ 1eV and in an insulator is ≥ 3eV. E1=0, E2=1eV, E3≥3eV . E1 < E2 < E3. 3. Find the truth table of following gates….4. Find the truth table of following gates….5.Name the fundamental gates. Ans : OR, AND and NOT gates. 6.What is the width of dipletion layer in p-n junction diode? Ans : The depletion layer in junction diode is of the order of micrometer (≈10-6m) 7.The current gain (α) of a transister in common base configuration is 0.98. What does It physically mean? Ans :The current gain α=0.98 means that 98% of charge carriers of an emitter reach the collector and constitute the collector current 8.In the following diagrams, write which of the diodes are forward biased and which are reverse biased Ans : The junction diode in figs (b) and (d) are forward biased and (a) and (c) are reverse biased. 9.Name the gate obtained from the combination of gates shown if figure. Draw the