S.No Questions Answers Q.1 What are permanent magnets? Give one example. Ans.1 A magnet that retains its magnetic properties in the absence of an inducing field or current is known as permanent magnet. Permanent magnets are those magnets which have high retentivity and coercivity. For example: steel. Q.2 What is the geometrical shape of equipotential surfaces due to a single isolated charge? Ans.2 The equipotential surfaces due to a single isolated charge are concentric spherical shells and the distance between the shells increases with the decrease. Q.3 Which of the following waves can be polarized (i) Heat waves (ii) Sound waves? Give reason to support your answer. Q.4 A capacitor has been charged by a dc source. What are the magnitude of conduction and displacement current, when it is fully charged? Q.5 Write the relationship between angle of incidence i, angle of prism A and angle of minimum deviations for a triangular prism. Ans.3 Polarization is the property of transverse waves. (i)heat waves are transverse in nature (Transverse waves can oscillate in the direction perpendicular to the direction of its propagation) so heat waves can be polarized. (ii) Sound waves cannot be polarized because sound waves are longitudinal in nature (longitudinal waves oscillate only along the direction of its propagation). Ans.4 Electric flux through the plates of capacitor is given by φ E = Here, q is constant, when capacitor is fully charged. Displacement current, I D = ε 0 = 0 Conduction current, I= C = 0 (as Voltage becomes constant when capacitor is fully charged) Ans.5 The relationship between angle of incidence i, angle of prism A and angle of minimum deviations for a triangular prism is Q.6 The given graph shows the variation of photoelectric current (I) versus applied voltage (V) for two difference photosensitive materials and for two different intensities of the incident radiations. Identify the pairs of curves that correspond to different materials but same intensity of incident radiation. Ans.6
Q.7 A 10 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 38Ω as shown in the figure. Find the value of the current in circuit. The pairs of curves (1 and 3) and (2 and 4) correspond to different materials but same intensity of incident radiation. Ans.7 Q.8 The emf of a cell is always greater than its terminal voltage. Why? Give reason. Q.9 (a) Write the necessary conditions for the phenomenon of total internal reflection to occur. (b) Write the relation between the refractive index and critical angle for a given pair of optical media. Q.10 State Lenz s Law. A metallic rod held horizontally along east west direction, is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer. Q.11 A convex lens of focal length 25 cm is placed coaxially in contact with a concave lens of focal length 20 cm. Determine the power of the combination. Will the system be converging or diverging in nature? The equivalent emf of the batteries is given by ε= 200-10= 190V So, the current in the circuit is given by I= = = 5 A Ans.8 Emf is the potential difference across the two terminals of a voltage source when it is not connected to any circuit. But when it is connected to a circuit, the voltage reduces slightly because of the internal resistance of the voltage source. The terminal voltage of the voltage source is the potential difference across the terminal when it is connected to a circuit. This is why the emf of a cell is always greater than its terminal voltage. Ans.9 (a) The necessary conditions for the phenomenon of total internal reflection to occur are: (i) the light is in the more dense medium and approaching the less dense medium. (ii) The angle of incidence is greater than the so-called critical angle. (b) Then the refractive index of the second medium is: n = sin i /sin r = sin 90 / sin c => n = 1 / sin c So, REFRACTIVE INDEX is n = 1 / sin c where, c = critical angle for the medium Ans.10 Lenz s law stats that, An induced electromotive force (emf) always gives rise to a current whose magnetic field oppose the original change in magnetic flux. Yes, emf will be induced in the rod as there is change in magnetic flux. When a metallic rod held horizontally along eastwest direction, is allowed to fall freely under gravity i.e. fall from north to south, the intensity of magnetic lines of earth s magnetic field changes through it i.e. the magnetic flux changes and hence the induced emf in it. Ans.11 Focal length of convex lens Focal length of concave lens Equivalent focal length
Power of the combination Q.12 An ammeter of resistance 0.80 Ω can measure current up to 1.0A. (i) What must be the value of shunt resistance to enable the ammeter to measure current up to 5.0 A? (ii) What is the combined resistance of the ammeter and the shunt? The focal length of the combination = 1 m = 100 cm. As the focal length is in negative, the system will be diverging in nature. Ans.12 Given: R A = 0.80 ohm and I A = 1.0 A. So, voltage across ammeter, V= IR = 1.0X 0.80= 0.8 V Let the value of shunt = x (i) Resistance of ammeter with shunt I= 5A, V= 0.8 V Using, R = =. Therefore, the shunt resistance is 0.2 ohm. (ii) Combined resistance of the ammeter and the shunt, Q.13. In the given circuit diagram a voltmeter V is connected across a lamp L. How would (i) the brightness of the lamp and (ii) voltmeter reading V be affected, if the value of resistance R is decreased? Justify your answer. Ans.13 The given circuit diagram is Common Emitter (CE) configuration of an npntransistor. The input circuit is forward biased and collector circuit is reverse biased. (i)if the base resistance R decreases, the input circuit will become more forward biased thus base current (IB) decreases and emitter current (IE) increases. This will increase the collector current (IC) as IE = IB + IC. When IC increases which flows through the lamp, the voltage across the bulb will also increase thus brightness of the lamp increases. (ii)as the voltmeter is connected in parallel with the lamp, the reading in the voltmeter will also increase. Q.14 (a) An em wave is travelling in a medium with a velocity = v ı. Ans.14 (a) Given that velocity, V = v î and electric field, E along X-axis and magnetic field,
Draw a sketch showing the propagation of the em wave, indicating the direction of the oscillating electric and magnetic fields. (b) How are the magnitudes of the electric and magnetic fields related to velocity of the em wave? B along Z-axis. The propagation of EM wave is shown below: (b) The magnitudes of the electric and magnetic fields related to velocity of the em wave Q.15 Block diagram of a receiver is shown in the figure: (a) Identify X and Y. (b) Write their functions. Q.16 Explain, with the help of a circuit diagram, the working of a photodiode. Write briefly how it is used to detect the optical signals. Mention the important considerations required while fabricating a pn junction diode to be used as a Light Emitting Diode (LED). What should be the order of band gap of an LED if it is required to emit light in the visible range? according to the relation given: Ans.15 (a) X represents Intermediate Frequency (IF) stage and Y represents an Amplifier. (b) At IF stage, the carrier frequency is changed to a lower frequency and in this process, the modulated signal is detected. This is the function of IF stage. While the function of amplifier is to amplify the detected signal. Ans.16 A photodiode is a semiconductor device that converts light into current. The current is generated when photons are absorbed in the photodiode. Working: A photodiode is a p-n junction diode whose function is controlled by the light allowed to fall on it. Suppose, the wavelength is such that the energy of a photon, hc/λ, is sufficient to break a valance bond. When such light falls on the junction, new electron-hole pairs are created. The number of charge carriers increases and hence the conductivity of the junction increases. If the junction is connected in some circuit, the current in the circuit is controlled by the intensity of the incident light. The important considerations required while fabricating a p-n junction diode to be used as a Light Emitting Diode are: (i)the reverse breakdown voltages of LEDs are very low, typically around 5V. So care should be taken while fabricating a p-n junction diode so that the high reverse voltages do not appear across them.
Q.17 Write three important factors which justify the need of modulating a message signal. Show diagrammatically how an amplitude modulated wave is obtained when a modulating signal is superimposed on a carrier wave. (ii) There is very little resistance to limit the current in LED. Therefore, a resistor must be used in series with the LED to avoid any damage to it. The semiconductor used for fabrication of visible LEDs must at least have a band gap of 1.8 ev (spectral range of visible light is from about 0.4 µm to 0.7 µm, i.e., from about 3 ev to 1.8 ev). Ans.17 (i)size of antenna or aerial: For communication within the effective length of the antennas, the transmitting frequencies should be high, so modulation is required. (ii) Effective power which is radiated by antenna: Since the power radiated by an antenna of length l is proportional to (l/λ) 2. As high powers are needed for good transmission so, higher frequency is required which can be achieved by modulation. (iii) The interference of signals from different transmitters: To avoid the interference of the signals there is need of high frequency which can be achieved by the modulation. In order to separate the various signals, radio stations must broadcast at different frequencies. Q.18 A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 360 µc. When potential across the capacitor is reduced by 120 V, the charge stored in it becomes 120 µc. Calculate: (i) The potential V and the unknown capacitance C. (ii) What will be the charge stored in the capacitor, if the voltage applied had increased by 120 V? A hollow cylindrical box of length 1 m and area of cross section25 cm2 is placed in a three dimensional coordinate system as shown in the figure. The electric field in the region is given by = 50 x ı, where E is NC 1 and x is in meters. Find (i) Net flux through the cylinder. (ii) Charge enclosed by the cylinder. Ans.18 (i) Initial voltage, V1 = V volts and charge stored, Q1 = 360 µc. Changed potential, V2 = V 120 and Q2 = 120 µc Q1 = CV1 (1) Q2 = CV2...(2) By dividing (2) from (1), we get If the voltage applied had increased by 120 V, (i) then V3 = 180 + 120 = 300 V
Hence, charge stored in the capacitor, Q3 = CV3 = 2 106 300= 600 µc (i) Q.19 (a) In a typical nuclear reaction, e.g. although number of nucleons is conserved, yet energy is released. How? Explain. (b) Show that nuclear density in a given nucleus is independent of mass number Ans.19 (a) In a nuclear reaction, the sum of the masses of the target nucleus bombarding particle may be greater or less than the sum of the masses of the and the product nucleus and the outgoing particle. So from the law of conservation of mass energy some energy (3.27 MeV) is evolved or involved in a nuclear reaction. This energy is called Q-value of the nuclear reaction. (b) Density of the nucleus= Q.20 (a) Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons. (R 0 = 1.1 X 10-15 m) which shows that the density is independent of mass number A. Ans.20 (a) Photoelectric effect cannot be explained on the basis of wave nature of light because Classical wave theory cannot explain the first 3 observations of photoelectric
(b) Write the basic features of photon picture of electromagnetic radiation on which Einstein s photoelectric equation is based. Q.21 A metallic rod of length l is rotated with a frequency v with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius r, about an axis passing through the centre and perpendicular to the plane of the ring. A constant uniform magnetic field B parallel to the axis is present everywhere. Using Lorentz force, explain how emf is induced between the centre and the metallic ring and hence obtained the expression for it. effect: 1. Existence of the threshold frequency Since energy of the wave is dependent on the square of its amplitude, the classical wave theory predicts that if sufficiently intense light is used, the electrons would absorb enough energy to escape. There should not be any threshold frequency. 2. Almost immediate emission of photoelectrons Based on classical wave theory, electrons require a period of time before sufficient energy is absorbed for it to escape from the metal. Accordingly, a dim light after some delay would transfer sufficient energy to the electrons for ejection, whereas a very bright light would eject electrons after a short while. However, this did not happen in photoelectric effect. The independence of kinetic energy of photoelectron on intensity and the dependence on frequency According to classical wave theory, if light of higher intensity is used, the kinetic energy of an ejected electron can be increased. This is because the greater the intensity, the larger the energy of the light wave striking the metal surface, so electrons are ejected with greater kinetic energy. However, it cannot explain why maximum kinetic energy is dependent on the frequency and independent of intensity (b) Photon picture of electromagnetic radiation on which Einstein s photoelectric equation is based on particle nature of light. Its basic features are: (i) In interaction with matter, radiation behaves as if it is made up of particles called photons. (ii) Each photon has energy E (=hν) and momentum p (=hν/c), and speed c, the speed of light. (iii) All photons of light of a particular frequency ν, or wavelength λ, have the same energy E (=hν=hc/λ) and momentum p (=hν/c=h/λ), whatever the intensity of radiation may be. (iv) By increasing the intensity of light of given wavelength, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy. Thus, photon energy is independent of intensity of radiation. (v) Photons are electrically neutral and are not deflected by electric and magnetic fields. (vi) In a photonparticle collision (such as photonelectron collision), the total energy and total momentum are conserved. However, number of photons may not be conserved. Ans.21 Suppose the length of the rod is greater than the radius of the circle and rod rotates anticlockwise and suppose the direction of electrons in the rod at any instant be along +y direction. Suppose the direction of the magnetic field is along +z direction. Then, using Lorentz law,
Thus, the direction of force on the electrons is along x axis. so, the electrons will move towards the center i.e., the fixed end of the rod. This movement of electrons will result in current and hence it will produce emf in the rod between the fixed end and the point touching the ring. Let θ be the angle between the rod and radius of the circle at any time t. Q.22 Output characteristics of an npn transistor in CE configuration is shown in the figure. Determine: (i) dynamic output resistance (ii) dc current gain and (iii) ac current gain at an operating point VCE = 10 V, when IB = 30 µa. Ans.22 Q.23 Using Bohr s postulates, obtain the expression for the total energy of the electron in the stationary states of the hydrogen atom. Hence draw the energy level diagram showing how the line spectra corresponding to Balmer series occur due to transition between energy levels. Ans.23 According to Bohr s postulates, in a hydrogen atom, a single alectron revolves around a nucleus of charge +e. For an electron moving with a uniform speed in a circular orbit os a given radius, the centripetal force is provided by Columb force of attraction between the electron and the nucleus. The electron is held in a circular orbit by electrostatic attraction. The centripetal force is equal to the Coulomb force.so => ----------- (1) where m = mass of electron r = radius of electronic orbit v = velocity of electron. From (1), we have,
=> ----------- (2) (i) Kinetic energy of electron, (ii)potential energy From (2), we get Now, total energy of the electron in the nth orbit When the electron in a hydrogen atom jumps from higher energy level to the lower energy level, the difference of energies of the two energy levels is emitted as a radiation of particular wavelength. It is called a spectral line. In H-atom, when an electron jumps from the orbit ni to orbit nf, the wavelength of the emitted radiation is given by, Where, R Rydberg s constant = 1.09678 107 m 1 For Balmer series, nf = 2 and ni = 3, 4, 5, Where, ni = 3, 4, 5, These spectral lines lie in the visible region.
Q.24 (a) In what way is diffraction from each slit related to the interference pattern in a double slit experiment. (b) Two wavelengths of sodium light 590 nm and 596 nm are used, in turn to study the diffraction taking place at a single slit of aperture 2 10 4 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases. Ans.24 (a) If the width of each slit is comparable to the wavelength of light used, the interference pattern thus obtained in the double slit experiment is modified by diffraction from each of the two slits. (b) given: λ1 = 590 nm = 5.9 10 7 m λ2 = 596 nm = 5.96 10 7 m Distance of the slits from the screen D = 1.5 m Distance between the two slits = a = 2 104m For the first secondary maxima, separation between the positions of the first secondary maxima of two spectral lines, Q.25 In a series LCR circuit connected to an ac source of variable frequency and voltage ν = vm sin ωt, draw a plot showing the variation of current (I) with angular frequency (ω) for two different values of resistance R1 and R2 (R1 > R2). Write the condition under which the phenomenon of resonance occurs. For which value of the resistance out of the two curves, a sharper resonance is produced? Define Q-factor of the circuit and give its significance. Ans.25 The condition under which the phenomenon of resonance occurs is The current amplitude is maximum at the resonant frequency ω. The variation of im with ω in a LCR series circuit for two values of Resistance R1 and R2 (R1 > R2), is shown below,
Since im = vm / R at resonance, the current amplitude for case R2 is sharper to that for case R1. Q factor of a resonant LCR circuit is defined as the ratio of voltage drop across the capacitor (or inductor) to that of applied voltage. Or in other words is defined in terms of the ratio of the energy stored in the oscillating resonator to the energy dissipated per cycle by damping processes. Q.26 White travelling back to his residence in the car, Dr. Pathak was caught up in a thunderstorm. It became very dark. He stopped driving the car and waited for thunderstorm to stop. Suddenly he noticed a child walking alone on the road. He asked the boy to come inside the car till the thunderstorm stopped. Dr. Pathak dropped the boy at his residence. The body insisted that Dr. Pathak should meet his parents. The parents expressed their gratitude to Dr. Pathak for his concern for safety of the child. Answer the following questions based on the above information: (a) Why is it safer to sit inside a car during a thunderstorm? (b) Which two values are displayed by Dr. Pathak in his action? (c) Which values are reflected in parents' response to Dr. Pathak? (d) Give an example of similar action on your part in the past from everyday life. Q.27 (a) Draw a ray diagram showing the image formation by a compound microscope. Hence obtained expression for total magnification when the image is formed at infinity. (b) Distinguish between myopia and hypermetropia. Show Significance: Q factor determines the sharpness of the resonance.the circuit is close to the resonance for a larger range Δω of frequencies and the tuning of the circuit will not be good. So, less sharp the resonance, less is the selectivity of the circuit while higher is the Q, sharper is the resonance curve and lesser will be the loss in energy of the circuit. Ans.26 (a) It is safer to be inside a car during thunderstorm because the car acts like a Faraday cage. The metal in the car will shield you from any external electric fields and thus prevent the lightning from traveling within the car. (b) Awareness and Humanity (c) Gratitude and obliged (d) I once came across to a situation where a puppy was struck in the middle of a busy road during rain and was not able go cross due to heavy flow, so I quickly rushed and helped him. Ans.27 (a)the compound microscope consists essentially of two or more double convex lenses fixed in the two extremities of a hollow cylinder. The lower lens (nearest to the object) is called the objective; the upper lens (nearest to the eye of the observer), the eyepiece.
diagrammatically how these defects can be corrected. (a) State Huygen's principle. Using this principle draw a diagram to show how a plane wave front incident at the interface of the two media gets refracted when it propagates from a rarer to a denser medium. Hence verify Snell's law of refraction. (b) When monochromatic light travels from a rarer to a denser medium, explain the following, giving reasons: (i) Is the frequency of reflected and refracted light same as the frequency of incident light? (ii) Does the decrease in speed imply a reduction in the energy carried by light wave? Magnifying power, when final image is at infinity: The magnification produced by the compound microscope is the product of the magnifications produced by the eyepiece and objective. M =M o X M e-----------------(1) Where, Me and M0 are the magnifying powers of the eyepiece and objective respectively. Magnification produced by the objective is given by: M o = size of image / size of object M o = h / h M o = L/ f 0 ------------(2) Where, h, h' are object and image heights respectively and f0 is the focal length of the objective Magnification produced by the eye piece is given by: M e = size of image / size of object L is the tube length i.e. the distance between the second focal point of the objective and the first focal point of the eyepiece. When the final image is at infinity, then M e = D/ f e -------------(3) Magnifying power of compound microscope, from (1),(2) and (3): (b) Myopia A person suffering from myopia can see only nearby objects clearly, but cannot see the objects beyond a certain distance clearly.
Hypermetropia A person suffering from hypermetropia can see distant objects clearly, but cannot see nearby objects. Correction for Myopia : In order to correct the eye for this defect, a concave lens of suitable focal length is placed close to the eye so that the parallel ray of light from an object at infinity after refraction through the lens appears to come from the far point P of the myopic eye. Correction for Hypermetropia To correct this defect, a convex lens of suitable focal length is placed close to the eye so that the rays of light from an object placed at the point N after refraction through the lens appear to come from the near point N of the hypermetropic eye. Huygens had a very important insight into the nature of wave propagation which is nowadays called Huygens' principle. When applied to the propagation of light waves, this principle states that:
Every point on a wave-front may be considered a source of secondary spherical wavelets which spread out in the forward direction at the speed of light. The new wave-front is the tangential surface to all of these secondary wavelets. Refraction On The Basis Of Wave Theory: Consider any point Q on the incident wave front. Suppose when disturbance from point P on incident wave front reaches point P on the refracted wave front, the disturbance from point Q reaches Q' on the refracting surface XY. Since P A represents the refracted wave front, the time taken by light to travel from a point on incident wave front to the corresponding point on refracted wave front should always be the same. Now, time taken by light to go from Q to Q' will be t= In right angled ΔAQK, QAK = i QK = AK sin i (ii) In right angled Substituting (ii) and (iii) in equation (i),
The rays from different points on the incident wave front will take the same time to reach the corresponding points on the refracted wave front i.e., t given by equation (iv) is independent of AK. It will happen so, if This is the Snell s law for refraction of light. Q.28 a) State the working principle of a potentiometer. With the help of the circuit diagram, explain how a potentiometer is used to compare the emf's of two primary cells. Obtain the required expression used for comparing the emfs. (b) Write two possible causes for one sided deflection in a potentiometer experiment. (a) State Kirchhoff's rules for an electric network. Using Kirchhoff's rules, obtain the balance condition in terms of the resistances of four arms of Wheatstone bridge. (b) In the meterbridge experimental set up, shown in the figure, the null point D is obtained at a distance of 40 cm from end A of the meterbridge wire. If a resistance of 10Ω is connected in series with R1, null point is obtained at AD = 60 cm. Calculate the values of R1 and R2. (b) (i) The frequency of reflected and refracted light remains same as the frequency of incident light because frequency only depends on the source of light. (ii) Since the frequency remains same, hence there is no reduction in energy. Ans.28 (a) This is a very basic instrument used for comparing emf two cells and for calibrating ammeter, voltmeter and watt-meter. The basic working principle of potentiometer is very very simple. When a constant current is passed through a wire of uniform area of crosssection, the potential drop across any portion of the wire is directly proportional to the length of that portion. E1, E2 are the emf of the two cells. 1, 2, 3 form a two way key. When 1 and 3 are connected, E1 is connected to the galvanometer (G). Jokey is moved to N1, which is at a distance l1 from A, to find the balancing length. Applying loop rule to AN1G31A, Φ l1 + 0 E1 = 0..(1) Where, Φ is the potential drop per unit length Similarly, for E2 balanced against l2 (AN2), Φ l2 + 0 E2 = 0..(2) From equations (1) and (2), Thus we can compare the emf s of any two sources. Generally, one of the cells is chosen as a standard cell whose emf is known to a high degree of accuracy. The emf of the other cell is then calculated from Eq. (3). (b) (i) The potential difference between the ends of the potentiometer wire or the emf of the cell connected in the main circuit may not be greater than the emf of the cells whose emf are to be compared.
2. The positive terminals of the cells and the battery used in the circuit might not be connected to the same end of the potentiometer wire. (a)kirchhoff's rules: Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction. Alternatively, Σ i =0 Loop rule: The Algebraic sum of changes in the potential around any closed loop involving resistors and cells in the loop is zero. Alternatively, Σ Δ V =0, where Δ V is the changes in potential Wheatstone bridge: Where R1, R2, R3, R4 are the four resistances. Galvanometer G has a current i g flowing through it. At balance condition ig=0 Applying junction rule at B I 2 =I 4 Applying junction rule at D I 1 =I 3 Applying loop rule to closed loop ADBA -I 1 R 1 +0+I 2 R 2 =0 I 1 /I 2 = R 1 /R 2 ----------(1) Applying loop rule to closed loop CBDC I 2 R 4 +0-I 1 R 3 =0 (I 3 =I 1, I 4 =I 2 ) I 1 /I 2 = R 4 /R 3 -----------(2) From eq (1) and (2) = This is the balance condition in terms of the resistances of four arms of Wheatstone bridge. (b) Let R be the resistence per unit length of the potential meter wire.
Q.29 (a) Derive the expression for the torque on a rectangular current carrying loop suspended in a uniform magnetic field. (b) A proton and a deuteron having equal momenta enter in a region of uniform magnetic field at right angle to the direction of the field. Depict their trajectories in the field. (a) A small compass needle of magnetic moment m is free to turn about an axis perpendicular to the direction of uniform magnetic field B. The moment of inertia of the needle about the axis is I. The needle is slightly disturbed from its stable position and then released. Prove that it executes simple harmonic motion. Hence deduce the expression for its time period. (b) A compass needle, free to turn in a vertical plane orients itself with its axis vertical at a certain place on the earth. Find out the values of (i) horizontal component of earth s magnetic field and (ii) angle of dip at the place. Ans.29 (a) consider a loop ABCD Case1:The rectangular loop is placed such that the uniform magnetic field B is in the plane of loop.no force is exerted by the magnetic field B on the arms AD and BC. Magnetic field B exerts a force F1 on arm AB F1= IbB ---------(1) Magnetic field B exerts a force F2 on arm CD F2= IbB=F1 ---------(2) Net force on the loop=0 the torque on a rectangular current carrying loop rotates the loop in anticlockwise direction. Torque τ=
= I(ab)B Τ= B/A Where A=ab= area of coil Case2: Plane of the loop is not along the magnetic field.but makes angle with it. Force on AB=F1 Force on CD=F2 F1 and F2 are equal and opposite. So they cancel each other. F1=F2=IbB (b) Lorentz force= F= Bqv sinθ Where θ=90 0 So, F= Bqv Therefore, particle will move in circular path. If m p =mass of proton m d = mass of deuteron and v p = velocity of proton v d = velocity of deuteron charge of proton= charge of deuteron given m p v d =m d v d
Thus, trajectory of both particles will be same. (a)the torque on the needle Τ=mXB Τ= mb sinθ Thus, in equilibrium Negative sign shows that the restoring torque is in opposite direction to deflecting torque. For small value of θ, sinθ=θ This represents simple harmonic motion. (b)(i)the horizontal component of earth s magnetic field B H = Bcosδ Substituting, δ= 90 0 So, B H =0 (ii)for a compass needle align vertical plane angle of dip at the place δ= 90 0