Explore comprehensive solutions for Class 12 Science Physics Chapter 8 – Electromagnetic Waves in the NCERT textbook. These solutions offer step-by-step explanations, making them highly favored among students for quick completion of homework and effective exam preparation. The solutions cover all the questions from Chapter 8, providing valuable assistance to Class 12 Science Physics students. Access these free resources to enhance your understanding and excel in your studies. Page No 285: Question 8.1: Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A. (a) Calculate the capacitance and the rate of charge of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain. ANSWER: Radius of each circular plate, r = 12 cm = 0.12 m Distance between the plates, d = 5 cm = 0.05 m Charging current, I = 0.15 A Permittivity of free space, = 8.85 × 10−12 C2 N−1 m−2 (a) Capacitance between the two plates is given by the relation, C  Where, A = Area of each plate  Charge on each plate, q = CV Where, V = Potential difference across the plates Differentiation on both sides with respect to time (t) gives: Therefore, the change in potential difference between the plates is 1.87 ×109 V/s. (b) The displacement current across the plates is the same as the conduction current. Hence, the displacement current, id is 0.15 A. (c) Yes Kirchhoff’s first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current. Page No 286: Question 8.2: A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s−1. (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. ANSWER: Radius of each circular plate, R = 6.0 cm = 0.06 m Capacitance of a parallel plate capacitor, C = 100 pF = 100 × 10−12 F Supply voltage, V = 230 V Angular frequency, ω = 300 rad s−1 (a) Rms value of conduction current, I  Where, XC = Capacitive reactance ∴ I = V × ωC = 230 × 300 × 100 × 10−12 = 6.9 × 10−6 A = 6.9 μA Hence, the rms value of conduction current is 6.9 μA. (b) Yes, conduction current is equal to displacement current. (c) Magnetic field is given as: B  Where, μ0 = Free space permeability  I0 = Maximum value of current = r = Distance between the plates from the axis = 3.0 cm = 0.03 m ∴B  = 1.63 × 10−11 T Hence, the magnetic field at that point is 1.63 × 10−11 T. Page No 286: Question 8.3: What physical quantity is the same for X-rays of wavelength 10−10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500 m? ANSWER: The speed of light (3 × 108 m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum. Page No 286: Question 8.4: A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength? ANSWER: The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x–y plane. They are mutually perpendicular. Frequency of the wave, ν = 30 MHz = 30 × 106 s−1 Speed of light in a vacuum, c = 3 × 108 m/s Wavelength of a wave is given as: Page No 286: Question 8.5: A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? ANSWER: A radio can tune to minimum frequency, ν1 = 7.5 MHz= 7.5 × 106 Hz Maximum frequency, ν2 = 12 MHz = 12 × 106 Hz Speed of light, c = 3 × 108 m/s Corresponding wavelength for ν1 can be calculated as: Corresponding wavelength for ν2 can be calculated as: Thus, the wavelength band of the radio is 40 m to 25 m. Page No 286: Question 8.6: A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator? ANSWER: The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 109 Hz. Page No 286: Question 8.7: The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave? ANSWER: Amplitude of magnetic field of an electromagnetic wave in a vacuum, B0 = 510 nT = 510 × 10−9 T Speed of light in a vacuum, c = 3 × 108 m/s Amplitude of electric field of the electromagnetic wave is given by the relation, E = cB0 = 3 × 108 × 510 × 10−9 = 153 N/C Therefore, the electric field part of the wave is 153 N/C. Page No 286: Question 8.8: Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is ν = 50.0 MHz. (a) Determine, B0, ω, k, and λ. (b) Find expressions for E and B. ANSWER: Electric field amplitude, E0 = 120 N/C Frequency of source, ν = 50.0 MHz = 50 × 106 Hz Speed of light, c = 3 × 108 m/s (a) Magnitude of magnetic field strength is given as: Angular frequency of source is given as: ω = 2πν = 2π × 50 × 106 = 3.14 × 108 rad/s Propagation constant is given as: Wavelength of wave is given as: (b) Suppose the wave is propagating in the positive x direction. Then, the electric field vector will be in the positive y direction and the magnetic field vector will be in the positive z direction. This is because all three vectors are mutually perpendicular. Equation of electric field vector is given as: And, magnetic field vector is given …

NCERT Solutions for Class 12 Science Physics Chapter 8 – Electromagnetic Waves Read More »

Explore the comprehensive NCERT solutions for Class 12 Science Physics Chapter 7 on Alternating Current, featuring easy-to-follow step-by-step explanations. Widely favored among Physics students, these solutions are a valuable resource for homework completion and exam preparation. All the questions and answers from Chapter 7 of the NCERT Book for Class 12 Science Physics are available here at no cost, providing students with a convenient tool for efficient learning. Page No 266: Question 7.1: A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle? ANSWER: Resistance of the resistor, R = 100 Ω Supply voltage, V = 220 V Frequency, ν = 50 Hz (a) The rms value of current in the circuit is given as: (b) The net power consumed over a full cycle is given as: P = VI = 220 × 2.2 = 484 W Page No 266: Question 7.2: (a) The peak voltage of an ac supply is 300 V. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. What is the peak current? ANSWER: (a) Peak voltage of the ac supply, V0 = 300 V Rms voltage is given as: (b) Therms value of current is given as: I = 10 A Now, peak current is given as: Page No 266: Question 7.3: A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit. ANSWER: Inductance of inductor, L = 44 mH = 44 × 10−3 H Supply voltage, V = 220 V Frequency, ν = 50 Hz Angular frequency, ω=  Inductive reactance, XL = ω L Rms value of current is given as: Hence, the rms value of current in the circuit is 15.92 A. Page No 266: Question 7.4: A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit. ANSWER: Capacitance of capacitor, C = 60 μF = 60 × 10−6 F Supply voltage, V = 110 V Frequency, ν = 60 Hz Angular frequency, ω=  Capacitive reactance  Rms value of current is given as: Hence, the rms value of current is 2.49 A. Page No 266: Question 7.5: In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer. ANSWER: In the inductive circuit, Rms value of current, I = 15.92 A Rms value of voltage, V = 220 V Hence, the net power absorbed can be obtained by the relation, P = VI cos Φ Where, Φ = Phase difference between V and I For a pure inductive circuit, the phase difference between alternating voltage and current is 90° i.e., Φ= 90°. Hence, P = 0 i.e., the net power is zero. In the capacitive circuit, Rms value of current, I = 2.49 A Rms value of voltage, V = 110 V Hence, the net power absorbed can ve obtained as: P = VI Cos Φ For a pure capacitive circuit, the phase difference between alternating voltage and current is 90° i.e., Φ= 90°. Hence, P = 0 i.e., the net power is zero. Page No 266: Question 7.6: Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0 H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit? ANSWER: Inductance, L = 2.0 H Capacitance, C = 32 μF = 32 × 10−6 F Resistance, R = 10 Ω Resonant frequency is given by the relation, Now, Q-value of the circuit is given as: Hence, the Q-Value of this circuit is 25. Page No 266: Question 7.7: A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit? ANSWER: Capacitance, C = 30μF = 30×10−6F Inductance, L = 27 mH = 27 × 10−3 H Angular frequency is given as: Hence, the angular frequency of free oscillations of the circuit is 1.11 × 103 rad/s. Page No 266: Question 7.8: Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time? ANSWER: Capacitance of the capacitor, C = 30 μF = 30×10−6 F Inductance of the inductor, L = 27 mH = 27 × 10−3 H Charge on the capacitor, Q = 6 mC = 6 × 10−3 C Total energy stored in the capacitor can be calculated by the relation, Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor. Page No 266: Question 7.9: A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle? ANSWER: At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit. Resistance, R = 20 Ω Inductance, L = 1.5 H Capacitance, C = 35 μF = 30 × 10−6 F AC supply voltage to the LCR circuit, V = 200 V Impedance of the circuit is given by the relation, At resonance, Current in the circuit can be calculated as: Hence, the average power transferred to the circuit in one complete cycle= VI = 200 × 10 = 2000 W. Page No 266: Question 7.10: A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor? [Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.] ANSWER: The range of frequency (ν) of a radio is 800 kHz to 1200 kHz. Lower tuning frequency, ν1 = 800 kHz = 800 × 103 Hz Upper tuning frequency, ν2 = 1200 kHz = 1200× 103 Hz Effective inductance of circuit L = 200 μH = 200 × 10−6 H Capacitance of variable capacitor for ν1 is given as: C1 Where, ω1 = Angular frequency for capacitor C1 Capacitance of variable capacitor for ν2, C2 Where, ω2 = Angular frequency for capacitor C2 Hence, the range of the variable capacitor is from 88.04 pFto 198.1pF. Page No 266: Question 7.11: Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, R = …

NCERT Solutions for Class 12 Science Physics Chapter 7 – Alternating Current Read More »

Explore the comprehensive NCERT Solutions for Class 12 Science Physics Chapter 6 on Electromagnetic Induction. These solutions offer clear, step-by-step explanations that are immensely popular among Class 12 Science students. Whether you’re working on your homework or gearing up for exams, the Physics Electromagnetic Induction Solutions can be a valuable resource for quick and effective preparation. All questions and answers from Chapter 6 of the NCERT Book for Class 12 Science Physics are available here for free, ensuring you have a reliable and accessible aid to support your studies. Page No 229: Question 6.1: Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f ). (a) (b) (c) (d) (e) (f) ANSWER: The direction of the induced current in a closed loop is given by Lenz’s law. The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively. Using Lenz’s rule, the direction of the induced current in the given situations can be predicted as follows: (a) The direction of the induced current is along qrpq. (b) The direction of the induced current is along prqp. (c) The direction of the induced current is along yzxy. (d) The direction of the induced current is along zyxz. (e) The direction of the induced current is along xryx. (f) No current is induced since the field lines are lying in the plane of the closed loop. Page No 230: Question 6.2: Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19: (a) A wire of irregular shape turning into a circular shape; (b) A circular loop being deformed into a narrow straight wire. ANSWER: According to Lenz’s law, the direction of the induced emf is such that it tends to produce a current that opposes the change in the magnetic flux that produced it. (a) When the shape of the wire changes, the flux piercing through the unit surface area increases. As a result, the induced current produces an opposing flux. Hence, the induced current flows along adcb. (b) When the shape of a circular loop is deformed into a narrow straight wire, the flux piercing the surface decreases. Hence, the induced current flows along  Page No 230: Question 6.3: A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing? ANSWER: Number of turns on the solenoid = 15 turns/cm = 1500 turns/m Number of turns per unit length, n = 1500 turns The solenoid has a small loop of area, A = 2.0 cm2 = 2 × 10−4 m2 Current carried by the solenoid changes from 2 A to 4 A. Change in current in the solenoid, di = 4 − 2 = 2 A Change in time, dt = 0.1 s Induced emf in the solenoid is given by Faraday’s law as: Where,  = Induced flux through the small loop = BA … (ii) B = Magnetic field =  μ0 = Permeability of free space = 4π×10−7 H/m Hence, equation (i) reduces to: Hence, the induced voltage in the loop is  Page No 230: Question 6.4: A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s−1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case? ANSWER: Length of the rectangular wire, l = 8 cm = 0.08 m Width of the rectangular wire, b = 2 cm = 0.02 m Hence, area of the rectangular loop, A = lb = 0.08 × 0.02 = 16 × 10−4 m2 Magnetic field strength, B = 0.3 T Velocity of the loop, v = 1 cm/s = 0.01 m/s (a) Emf developed in the loop is given as: e = Blv = 0.3 × 0.08 × 0.01 = 2.4 × 10−4 V Hence, the induced voltage is 2.4 × 10−4 V which lasts for 2 s. (b) Emf developed, e = Bbv = 0.3 × 0.02 × 0.01 = 0.6 × 10−4 V Hence, the induced voltage is 0.6 × 10−4 V which lasts for 8 s. Page No 230: Question 6.5: A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s−1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring. ANSWER: Length of the rod, l = 1 m Angular frequency,ω = 400 rad/s Magnetic field strength, B = 0.5 T One end of the rod has zero linear velocity, while the other end has a linear velocity of lω. Average linear velocity of the rod, Emf developed between the centre and the ring, Hence, the emf developed between the centre and the ring is 100 V. Page No 230: Question 6.6: A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s−1 in a uniform horizontal magnetic field of magnitude 3.0×10−2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from? ANSWER: Max induced emf = 0.603 V Average induced emf = 0 V Max current in the coil = 0.0603 A Average power loss = 0.018 W (Power comes from the external rotor) Radius of the circular coil, r = 8 cm = 0.08 m Area of the coil, A = πr2 = π × (0.08)2 m2 Number of turns on the coil, N = 20 Angular speed, ω = 50 rad/s Magnetic field strength, B = 3 × 10−2 T …

NCERT Solutions for Class 12 Science Physics Chapter 6 – Electromagnetic Induction Read More »

Discover the NCERT Solutions for Class 12 Science Physics Chapter 4, focusing on Moving Charges and Magnetism. These solutions offer straightforward, step-by-step explanations that have gained immense popularity among Class 12 Science students. Whether you’re working on assignments or gearing up for exams, the Physics Moving Charges and Magnetism Solutions prove to be a convenient resource for swift completion of your homework and effective exam preparation. You can access all the questions and answers from Chapter 4 of the NCERT Book for Class 12 Science Physics here for free, ensuring that you have a valuable and readily available tool to support your studies. Page No 200: Question 5.1: Answer the following questions regarding earth’s magnetism: (a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field. (b) The angle of dip at a location in southern India is about 18º. Would you expect a greater or smaller dip angle in Britain? (c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground? (d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole? (e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of magnetic moment 8 × 1022 J T−1 located at its centre. Check the order of magnitude of this number in some way. (f ) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all? ANSWER: (a) The three independent quantities conventionally used for specifying earth’s magnetic field are: (i) Magnetic declination, (ii) Angle of dip, and (iii) Horizontal component of earth’s magnetic field (b)The angle of dip at a point depends on how far the point is located with respect to the North Pole or the South Pole. The angle of dip would be greater in Britain (it is about 70°) than in southern India because the location of Britain on the globe is closer to the magnetic North Pole. (c)It is hypothetically considered that a huge bar magnet is dipped inside earth with its north pole near the geographic South Pole and its south pole near the geographic North Pole. Magnetic field lines emanate from a magnetic north pole and terminate at a magnetic south pole. Hence, in a map depicting earth’s magnetic field lines, the field lines at Melbourne, Australia would seem to come out of the ground. (d)If a compass is located on the geomagnetic North Pole or South Pole, then the compass will be free to move in the horizontal plane while earth’s field is exactly vertical to the magnetic poles. In such a case, the compass can point in any direction. (e)Magnetic moment, M = 8 × 1022 J T−1 Radius of earth, r = 6.4 × 106 m Magnetic field strength,  Where, = Permeability of free space =  This quantity is of the order of magnitude of the observed field on earth. (f)Yes, there are several local poles on earth’s surface oriented in different directions. A magnetised mineral deposit is an example of a local N-S pole. Page No 200: Question 5.2: Answer the following questions: (a) The earth’s magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably? (b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why? (c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to sustain these currents? (d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past? (e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion? (f ) Interstellar space has an extremely weak magnetic field of the order of 10−12 T. Can such a weak field be of any significant consequence? Explain. [Note: Exercise 5.2 is meant mainly to arouse your curiosity. Answers to some questions above are tentative or unknown. Brief answers wherever possible are given at the end. For details, you should consult a good text on geomagnetism.] ANSWER: (a) Earth’s magnetic field changes with time. It takes a few hundred years to change by an appreciable amount. The variation in earth’s magnetic field with the time cannot be neglected. (b)Earth’s core contains molten iron. This form of iron is not ferromagnetic. Hence, this is not considered as a source of earth’s magnetism. (c)Theradioactivity in earth’s interior is the source of energy that sustains the currents in the outer conducting regions of earth’s core. These charged currents are considered to be responsible for earth’s magnetism. (d)Earth reversed the direction of its field several times during its history of 4 to 5 billion years. These magnetic fields got weakly recorded in rocks during their solidification. One can get clues about the geomagnetic history from the analysis of this rock magnetism. (e)Earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km) because of the presence of the ionosphere. In this region, earth’s field gets modified because of the field of single ions. While in motion, these ions produce the magnetic field associated with them. (f)An extremely weak magnetic field can bend charged particles moving in a circle. This may not be noticeable for a large radius path. With reference to the gigantic interstellar space, the deflection can affect the passage of charged particles. Page No 200: Question 5.3: A short bar magnet placed with its axis at 30º with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10−2 J. What …

NCERT Solutions for Class 12 Science Physics Chapter 5 – Magnetism And Matter Read More »

Explore the comprehensive NCERT Solutions for Class 12 Science Physics Chapter 4 – Moving Charges And Magnetism, featuring easy-to-follow, step-by-step explanations. These meticulously crafted solutions have gained immense popularity among Class 12 Science students, proving invaluable for completing homework assignments efficiently and preparing for exams. The Physics Moving Charges And Magnetism Solutions provided here are not only accessible but also serve as a handy resource for quick reference. All questions and answers from Chapter 4 of the NCERT Book for Class 12 Science Physics are available at your fingertips, ensuring that you have free and convenient access to the essential materials required for academic success. Page No 169: Question 4.1: A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil? ANSWER: Number of turns on the circular coil, n = 100 Radius of each turn, r = 8.0 cm = 0.08 m Current flowing in the coil, I = 0.4 A Magnitude of the magnetic field at the centre of the coil is given by the relation, Where,  = Permeability of free space = 4π × 10–7 T m A–1 Hence, the magnitude of the magnetic field is 3.14 × 10–4 T. Page No 169: Question 4.2: A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire? ANSWER: Current in the wire, I = 35 A Distance of a point from the wire, r = 20 cm = 0.2 m Magnitude of the magnetic field at this point is given as: B Where,  = Permeability of free space = 4π × 10–7 T m A–1 Hence, the magnitude of the magnetic field at a point 20 cm from the wire is 3.5 × 10–5 T. Page No 169: Question 4.3: A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire. ANSWER: Current in the wire, I = 50 A A point is 2.5 m away from the East of the wire.  Magnitude of the distance of the point from the wire, r = 2.5 m. Magnitude of the magnetic field at that point is given by the relation, B Where,  = Permeability of free space = 4π × 10–7 T m A–1 The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to the Maxwell’s right hand thumb rule, the direction of the magnetic field at the given point is vertically upward. Page No 169: Question 4.4: A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line? ANSWER: Current in the power line, I = 90 A Point is located below the power line at distance, r = 1.5 m Hence, magnetic field at that point is given by the relation, Where,  = Permeability of free space = 4π × 10–7 T m A–1 The current is flowing from East to West. The point is below the power line. Hence, according to Maxwell’s right hand thumb rule, the direction of the magnetic field is towards the South. Page No 169: Question 4.5: What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T? ANSWER: Current in the wire, I = 8 A Magnitude of the uniform magnetic field, B = 0.15 T Angle between the wire and magnetic field, θ = 30°. Magnetic force per unit length on the wire is given as: f = BI sinθ = 0.15 × 8 ×1 × sin30° = 0.6 N m–1 Hence, the magnetic force per unit length on the wire is 0.6 N m–1. Page No 169: Question 4.6: A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire? ANSWER: Length of the wire, l = 3 cm = 0.03 m Current flowing in the wire, I = 10 A Magnetic field, B = 0.27 T Angle between the current and magnetic field, θ = 90° Magnetic force exerted on the wire is given as: F = BIlsinθ = 0.27 × 10 × 0.03 sin90° = 8.1 × 10–2 N Hence, the magnetic force on the wire is 8.1 × 10–2 N. The direction of the force can be obtained from Fleming’s left hand rule. Page No 169: Question 4.7: Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A. ANSWER: Current flowing in wire A, IA = 8.0 A Current flowing in wire B, IB = 5.0 A Distance between the two wires, r = 4.0 cm = 0.04 m Length of a section of wire A, l = 10 cm = 0.1 m Force exerted on length l due to the magnetic field is given as: Where,  = Permeability of free space = 4π × 10–7 T m A–1 The magnitude of force is 2 × 10–5 N. This is an attractive force normal to A towards B because the direction of the currents in the wires is the same. Page No 169: Question 4.8: A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre. ANSWER: Length of the solenoid, l = 80 cm = 0.8 m There are five layers of windings of 400 turns each on the solenoid. Total number of turns on the solenoid, N = 5 × 400 = 2000 Diameter of the solenoid, D = 1.8 cm = 0.018 m Current carried by …

NCERT Solutions for Class 12 Science Physics Chapter 4 – Moving Charges And Magnetism Read More »

Explore comprehensive NCERT solutions for Class 12 Science Physics Chapter 3 on Current Electricity. These solutions offer clear and concise step-by-step explanations, making them highly favored by Physics students. Whether you need assistance with homework or are gearing up for exams, these solutions prove invaluable. All questions and answers from Chapter 3 of the NCERT Book for Class 12 Science Physics are provided here at no cost, ensuring you have easy access to reliable study material. Page No 127: Question 3.1: The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery? ANSWER: Emf of the battery, E = 12 V Internal resistance of the battery, r = 0.4 Ω Maximum current drawn from the battery = I According to Ohm’s law, The maximum current drawn from the given battery is 30 A. Page No 127: Question 3.2: A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed? ANSWER: Emf of the battery, E = 10 V Internal resistance of the battery, r = 3 Ω Current in the circuit, I = 0.5 A Resistance of the resistor = R The relation for current using Ohm’s law is, Terminal voltage of the resistor = V According to Ohm’s law, V = IR = 0.5 × 17 = 8.5 V Therefore, the resistance of the resistor is 17 Ω and the terminal voltage is 8.5 V. Page No 127: Question 3.3: (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor. ANSWER: (a) Three resistors of resistances 1 Ω, 2 Ω, and 3 Ω are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances. Total resistance = 1 + 2 + 3 = 6 Ω (b) Current flowing through the circuit = I Emf of the battery, E = 12 V Total resistance of the circuit, R = 6 Ω The relation for current using Ohm’s law is, Potential drop across 1 Ω resistor = V1 From Ohm’s law, the value of V1 can be obtained as V1 = 2 × 1= 2 V … (i) Potential drop across 2 Ω resistor = V2 Again, from Ohm’s law, the value of V2 can be obtained as V2 = 2 × 2= 4 V … (ii) Potential drop across 3 Ω resistor = V3 Again, from Ohm’s law, the value of V3 can be obtained as V3 = 2 × 3= 6 V … (iii) Therefore, the potential drop across 1 Ω, 2 Ω, and 3 Ω resistors are 2 V, 4 V, and 6 V respectively. Page No 127: Question 3.4: (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery. ANSWER: (a) There are three resistors of resistances, R1 = 2 Ω, R2 = 4 Ω, and R3 = 5 Ω They are connected in parallel. Hence, total resistance (R) of the combination is given by, Therefore, total resistance of the combination is. (b) Emf of the battery, V = 20 V Current (I1) flowing through resistor R1 is given by, Current (I2) flowing through resistor R2 is given by, Current (I3) flowing through resistor R3 is given by, Total current, I = I1 + I2 + I3 = 10 + 5 + 4 = 19 A Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the total current is 19 A. Page No 127: Question 3.5: At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is  ANSWER: Room temperature, T = 27°C Resistance of the heating element at T, R = 100 Ω Let T1 is the increased temperature of the filament. Resistance of the heating element at T1, R1 = 117 Ω Temperature co-efficient of the material of the filament, Therefore, at 1027°C, the resistance of the element is 117Ω. Page No 127: Question 3.6: A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10−7 m2, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment? ANSWER: Length of the wire, l =15 m Area of cross-section of the wire, a = 6.0 × 10−7 m2 Resistance of the material of the wire, R = 5.0 Ω Resistivity of the material of the wire = ρ Resistance is related with the resistivity as Therefore, the resistivity of the material is 2 × 10−7 Ω m. Page No 127: Question 3.7: A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver. ANSWER: Temperature, T1 = 27.5°C Resistance of the silver wire at T1, R1 = 2.1 Ω Temperature, T2 = 100°C Resistance of the silver wire at T2, R2 = 2.7 Ω Temperature coefficient of silver = α It is related with temperature and resistance as Therefore, the temperature coefficient of silver is 0.0039°C−1. Page No 127: Question 3.8: Aheating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds toa steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10−4 °C −1. ANSWER: Supply voltage, V = 230 V Initial current drawn, I1 = 3.2 A Initial resistance = R1, which is given by the relation, Steady state value of the current, I2 = 2.8 A Resistance at the steady state = R2, which is given as …

NCERT Solutions for Class 12 Science Physics Chapter 3 – Current Electricity Read More »

Unlock the NCERT Solutions for Class 12 Science Physics Chapter 2: Electrostatic Potential and Capacitance with clear, step-by-step explanations. These solutions have gained immense popularity among Physics students for their simplicity and effectiveness. Ideal for completing homework swiftly and gearing up for exams, the Electrostatic Potential and Capacitance Solutions ensure comprehensive understanding. Access all questions and answers from the NCERT Book of Class 12 Science Physics Chapter 2 without any cost. Page No 87: Question 2.1: Two charges 5 × 10−8 C and −3 × 10−8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. ANSWER: There are two charges, Distance between the two charges, d = 16 cm = 0.16 m Consider a point P on the line joining the two charges, as shown in the given figure. r = Distance of point P from charge q1 Let the electric potential (V) at point P be zero. Potential at point P is the sum of potentials caused by charges q1 and q2 respectively. Where, = Permittivity of free space For V = 0, equation (i) reduces to Therefore, the potential is zero at a distance of 10 cm from the positive charge between the charges. Suppose point P is outside the system of two charges at a distance s from the negative charge, where potential is zero, as shown in the following figure. For this arrangement, potential is given by, For V = 0, equation (ii) reduces to Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges. Page No 87: Question 2.2: A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon. ANSWER: The given figure shows six equal amount of charges, q, at the vertices of a regular hexagon. Where, Charge, q = 5 µC = 5 × 10−6 C Side of the hexagon, l = AB = BC = CD = DE = EF = FA = 10 cm Distance of each vertex from centre O, d = 10 cm Electric potential at point O, Where, = Permittivity of free space Therefore, the potential at the centre of the hexagon is 2.7 × 106 V. Page No 87: Question 2.3: Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface? ANSWER: (a) The situation is represented in the given figure. An equipotential surface is the plane on which total potential is zero everywhere. This plane is normal to line AB. The plane is located at the mid-point of line AB because the magnitude of charges is the same. (b) The direction of the electric field at every point on this surface is normal to the plane in the direction of AB. Page No 87: Question 2.4: A spherical conductor of radius 12 cm has a charge of 1.6 × 10−7C distributed uniformly on its surface. What is the electric field (a) Inside the sphere (b) Just outside the sphere (c) At a point 18 cm from the centre of the sphere? ANSWER: (a) Radius of the spherical conductor, r = 12 cm = 0.12 m Charge is uniformly distributed over the conductor, q = 1.6 × 10−7 C Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it. (b) Electric field E just outside the conductor is given by the relation, Where, = Permittivity of free space Therefore, the electric field just outside the sphere is . (c) Electric field at a point 18 m from the centre of the sphere = E1 Distance of the point from the centre, d = 18 cm = 0.18 m Therefore, the electric field at a point 18 cm from the centre of the sphere is . Page No 87: Question 2.5: A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10−12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6? ANSWER: Capacitance between the parallel plates of the capacitor, C = 8 pF Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, k = 1 Capacitance, C, is given by the formula, Where, A = Area of each plate = Permittivity of free space If distance between the plates is reduced to half, then new distance, d’ =  Dielectric constant of the substance filled in between the plates,  = 6 Hence, capacitance of the capacitor becomes Taking ratios of equations (i) and (ii), we obtain Therefore, the capacitance between the plates is 96 pF. Page No 87: Question 2.6: Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply? ANSWER: (a) Capacitance of each of the three capacitors, C = 9 pF Equivalent capacitance (C’) of the combination of the capacitors is given by the relation, Therefore, total capacitance of the combination is 3 pF3 pF. (b) Supply voltage, V = 120 V Potential difference (V‘) across each capacitor is equal to one-third of the supply voltage. Therefore, the potential difference across each capacitor is 40 V. Page No 87: Question 2.7: Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply. ANSWER: (a) Capacitances of the given capacitors are For the parallel combination of the capacitors, equivalent capacitoris given by the algebraic sum, Therefore, total capacitance of the combination is 9 pF. (b) Supply voltage, V = 100 V The voltage through all the three capacitors is same = V = 100 V Charge on a capacitor of capacitance C and potential difference V is given by the relation, q = VC â€¦ (i) For C = 2 pF, …

NCERT Solutions for Class 12 Science Physics Chapter 2 – Electrostatic Potential And Capacitance Read More »

Explore the comprehensive NCERT Solutions for Class 12 Science Physics Chapter 1: Electric Charges and Fields, featuring straightforward, step-by-step explanations. These solutions have gained immense popularity among Class 12 Science students, proving invaluable for efficiently completing homework assignments and preparing for exams. The Physics Electric Charges and Fields Solutions offer a user-friendly resource to enhance your understanding of the subject. All questions and answers from Chapter 1 of the NCERT Book for Class 12 Science Physics are available here at no cost, ensuring easy access to valuable study material. Page No 46: Question 1.1: What is the force between two small charged spheres having charges of 2 × 10−7 C and 3 × 10−7 C placed 30 cm apart in air? ANSWER: Repulsive force of magnitude 6 × 10−3 N Charge on the first sphere, q1 = 2 × 10−7 C Charge on the second sphere, q2 = 3 × 10−7 C Distance between the spheres, r = 30 cm = 0.3 m Electrostatic force between the spheres is given by the relation, Where, ∈0 = Permittivity of free space Hence, force between the two small charged spheres is 6 × 10−3 N. The charges are of same nature. Hence, force between them will be repulsive. Page No 46: Question 1.2: The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge − 0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? ANSWER: (a) Electrostatic force on the first sphere, F = 0.2 N Charge on this sphere, q1 = 0.4 μC = 0.4 × 10−6 C Charge on the second sphere, q2= − 0.8 μC = − 0.8 × 10−6C Electrostatic force between the spheres is given by the relation, Where, ∈0 = Permittivity of free space The distance between the two spheres is 0.12 m. (b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N. Page No 46: Question 1.3: Check that the ratio ke2/G mempis dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify? ANSWER: The given ratio is . Where, G = Gravitational constant Its unit is N m2 kg−2. me and mp = Masses of electron and proton. Their unit is kg. e = Electric charge. Its unit is C. ∈0 = Permittivity of free space Its unit is N m2 C−2. Hence, the given ratio is dimensionless. e = 1.6 × 10−19 C G = 6.67 × 10−11 N m2 kg-2 me= 9.1 × 10−31 kg mp = 1.66 × 10−27 kg Hence, the numerical value of the given ratio is This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant. Page No 46: Question 1.4: (a) Explain the meaning of the statement ‘electric charge of a body is quantised’. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges? ANSWER: (a) Electric charge of a body is quantized. This means that only integral (1, 2, …., n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge. (b) In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous. Page No 46: Question 1.5: When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge. ANSWER: Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies. Page No 46: Question 1.6: Four point charges qA = 2 μC, qB = −5 μC, qC = 2 μC, and qD = −5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square? ANSWER: The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square. Where, (Sides) AB = BC = CD = AD = 10 cm (Diagonals) AC = BD = cm AO = OC = DO = OB = cm A charge of amount 1μC is placed at point O. Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1 μC charge at centre O is zero. Page No 46: Question 1.7: (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not? (b) Explain why two field lines never cross each other at any point? ANSWER: (a) An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the …

NCERT Solutions for Class 12 Science Physics Chapter 1 – Electric Charges And Fields Read More »