Q11.The number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V is : ( Given e = 1.6 × 10−19C) (1) 6.25 × 1017 (2) 1.25 × 1019 (3) 6.25 × 1018 (4) 31.25 × 1017 JEE Main 2024 (06 Apr Shift 2) JEE Main Previous Year Paper
What This Question Tests
This question requires calculating current from power and voltage, and then using the quantization of charge to find the number of electrons flowing per second.
Concepts Tested
Formulas Used
P = VI
I = Q/t
Q = ne
📚 NCERT Sections This Tests
3.3 — At Room Temperature (27.0 °C) The Resistance Of A Heating Element
Physics Class 11 · Chapter 3
3.3 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 1.70 × 10–4 °C–1. 105 Reprint 2025-26 Physics 3.4 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 W. What is the resistivity of the material at the temperature of the experiment? 3.5 A silver wire has a resistance of 2.1 W at 27.5 °C, and a resistance of 2.7 W at 100 °C. Determine the temperature coefficient of resistivity of silver. 3.6 A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a 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. 3.7 Determine the current in each branch of the network shown in Fig. 3.20: FIGURE 3.20 3.8 A storage battery of emf 8.0 V and internal resistance 0.5 W is being charged by a 120 V dc supply using a series resistor of 15.5 W. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit? 3.9 The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 1028 m–3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10–6 m2 and it is carrying a current of 3.0 A. Reprint 2025-26 Chapter Four MOVING CHARGES AND MAGNETISM 4.1 INTRODUCTION Both Electricity and Magnetism have been known for more than 2000 years. However, it was only about 200 years ago, in 1820, that it was realised that they were intimately related. During a lecture demonstration in the summer of 1820, Danish physicist Hans Christian Oersted noticed that a current in a straight wire caused a noticeable deflection in a nearby magnetic compass needle. He investigated this phenomenon. He found that the alignment of the needle is tangential to an imaginary circle which has the straight wire as its centre and has its plane perpendicular to the wire. This situation is depicted in Fig.4.1(a). It is noticeable when the current is large and the needle sufficiently close to the wire so that the earth’s magnetic field may be ignored. Reversing the direction of the current reverses the orientation of the needle [Fig. 4.1(b)]. The deflection increases on increasing the current or bringing the needle closer to the wire. Iron filings sprinkled around the wire arrange themselves in concentric circles with the wire as the centre [Fig. 4.1(c)]. Oersted concluded that moving charges or currents produced a magnetic field in the surrounding space. Following this, there was intense experimentation. In 1864, the laws obeyed by electricity and magnetism were unified and formulated by Reprint 2025-26 Physics James Maxwell who then realised that light was electromagnetic waves. Radio waves were discovered by Hertz, and produced by J.C.Bose and G. Marconi by the end of the 19th century. A remarkable scientific and technological progress took place in the 20th century. This was due to our increased understanding of electromagnetism and the invention of devices for production, amplification, transmission and detection of electromagnetic waves. FIGURE 4.1 The magnetic field due to a straight long current-carrying wire. The wire is perpendicular to the plane of the paper. A ring of compass needles surrounds the wire. The orientation of the needles is shown when (a) the current emerges out of the plane of the paper, (b) the current moves into the plane of the paper. (c) The arrangement of iron filings around the wire. The darkened ends of the needle represent north poles. The effect of the earth’s magnetic field is neglected. In this chapter, we will see how magnetic field exerts forces on moving charged particles, like electrons, protons, and current-carrying wires. We shall also learn how currents produce magnetic fields. We shall see how particles can be accelerated to very high energies in a cyclotron. We shall study how currents and voltages are detected by a galvanometer.(1777–1851) In this and subsequent Chapter on magnetism, we adopt the following convention: A current or a field (electric or magnetic) emerging out of the plane of the paper is depicted by a dot (¤). A current or a field going into the plane of the paper is depicted by a cross ()*. Hans Christian Oersted Figures. 4.1(a) and 4.1(b) correspond to these twoOERSTED (1777–1851) Danish situations, respectively. physicist and chemist, professor at Copenhagen. 4.2 MAGNETIC FORCE He observed that a compass needle suffers a 4.2.1 Sources and fields deflection when placed Before we introduce the concept of a magnetic field B, weCHRISTIAN near a wire carrying an electric current. This shall recapitulate what we have learnt in Chapter 1 about discovery gave the first the electric field E. We have seen that the interaction empirical evidence of a between two charges can be considered in two stages.HANS connection between electric The charge Q, the source of the field, produces an electric and magnetic phenomena. field E, where * A dot appears like the tip of an arrow pointed at you, a cross is like the feathered 108 tail of an arrow moving away from you. Reprint 2025-26 Moving Charges and Magnetism E = Q ˆr / (4pe0)r2 (4.1) where ˆr is unit vector along r, and the field E is a vector field. A charge q interacts with this field and experiences a force F given by F = q E = q Q ˆr / (4pe0) r 2 (4.2) As pointed out in the Chapter 1, the field E is not just an artefact but has a physical role. It can convey energy and momentum and is not established instantaneously but takes finite time to propagate. The concept of a field was specially stressed by Faraday and was incorporated by Maxwell in his unification of electricity and magnetism. In addition to depending on each point in space, it can also vary with time, i.e., be a function of time. In our Hendrik Antoon Lorentz discussions in this chapter, we will assume that the fields (1853 – 1928) Dutch do not change with time. theoretical physicist, The field at a particular point can be due to one or professor at Leiden. He investigated themore charges. If there are more charges the fields add HENDRIK relationship between vectorially. You have already learnt in Chapter 1 that this electricity, magnetism, and is called the principle of superposition. Once the field is mechanics. In order to known, the force on a test charge is given by Eq. (4.2). explain the observed effect Just as static charges produce an electric field, the of magnetic fields on ANTOONcurrents or moving charges produce (in addition) a emitters of light (Zeeman magnetic field, denoted by B (r), again a vector field. It effect), he postulated the existence of electric chargeshas several basic properties identical to the electric field. in the atom, for which he It is defined at each point in space (and can in addition was awarded the Nobel Prize depend on time). Experimentally, it is found to obey the in 1902. He derived a set of LORENTZprinciple of superposition: the magnetic field of several transformation equations sources is the vector addition of magnetic field of each (known after him, as individual source. Lorentz transformation equations) by some tangled (1853 4.2.2 Magnetic Field, Lorentz Force mathematical arguments, – but he was not aware thatLet us suppose that there is a point charge q (moving these equations hinge on a with a velocity v and, located at r at a given time t) in new concept of space andpresence of both the electric field E (r) and the magnetic 1928) time. field B (r). The force on an electric charge q due to both of them can be written as F = q [ E (r) + v × B (r)] º Felectric +Fmagnetic (4.3) This force was given first by H.A. Lorentz based on the extensive experiments of Ampere and others. It is called the Lorentz force. You have already studied in detail the force due to the electric field. If we look at the interaction with the magnetic field, we find the following features. (i) It depends on q, v and B (charge of the particle, the velocity and the magnetic field). Force on a negative charge is opposite to that on a positive charge. (ii) The magnetic force q [ v × B ] includes a vector product of velocity 109 and magnetic field. The vector product makes the force due to magnetic Reprint 2025-26 Physics field vanish (become zero) if velocity and magnetic field are parallel or anti-parallel. The force acts in a (sideways) direction perpendicular to both the velocity and the magnetic field. Its direction is given by the screw rule or right hand rule for vector (or cross) product as illustrated in Fig. 4.2. (iii) The magnetic force is zero if charge is not moving (as then |v|= 0). Only a moving charge feels the magnetic force. The expression for the magnetic force helps us to define the unit of the magnetic field, if one FIGURE 4.2 The direction of the magnetic takes q, F and v, all to be unity in the force force acting on a charged particle. (a) The equation F = q [ v × B] =q v B sin q ˆn , where q is force on a positively charged particle with the angle between v and B [see Fig. 4.2 (a)]. The velocity v and making an angle q with the magnitude of magnetic field B is 1 SI unit, when magnetic field B is given by the right-hand the force acting on a unit charge (1 C), moving rule. (b) A moving charged particle q is perpendicular to B with a speed 1m/s, is one deflected in an opposite sense to –q in the newton. presence of magnetic field. Dimensionally, we have [B] = [F/qv] and the unit of B are Newton second / (coulomb metre). This unit is called tesla (T) named after Nikola Tesla (1856 – 1943). Tesla is a rather large unit. A smaller unit (non-SI) called gauss (=10–4 tesla) is also often used. The earth’s magnetic field is about 3.6 × 10–5 T. 4.2.3 Magnetic force on a current-carrying conductor We can extend the analysis for force due to magnetic field on a single moving charge to a straight rod carrying current. Consider a rod of a uniform cross-sectional area A and length l. We shall assume one kind of mobile carriers as in a conductor (here electrons). Let the number density of these mobile charge carriers in it be n. Then the total number of mobile charge carriers in it is nlA. For a steady current I in this conducting rod, we may assume that each mobile carrier has an average drift velocity vd (see Chapter 3). In the presence of an external magnetic field B, the force on these carriers is: F = (nlA)q vd ´ B where q is the value of the charge on a carrier. Now nq vd is the current density j and |(nq vd)|A is the current I (see Chapter 3 for the discussion of current and current density). Thus, F = [(nq vd )lA] × B = [ jAl ] ´ B = Il ´ B (4.4) where l is a vector of magnitude l, the length of the rod, and with a direction identical to the current I. Note that the current I is not a vector. In the last step leading to Eq. (4.4), we have transferred the vector sign from j to l. Equation (4.4) holds for a straight rod. In this equation, B is the external magnetic field. It is not the field produced by the current-carrying rod. If the wire has an arbitrary shape we can calculate the Lorentz force on it by considering it as a collection of linear strips dlj and summing F Idl j × B j This summation can be converted to an integral in most cases. Reprint 2025-26 Moving Charges and Magnetism Example 4.1 A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B (Fig. 4.3). What is the magnitude of the magnetic field? FIGURE 4.3 Solution From Eq. (4.4), we find that there is an upward force F, of magnitude IlB,. For mid-air suspension, this must be balanced by the force due to gravity: m g = I lB m g B = I l Interactive Charged 0.2 × 9.8 = = 0.65 T 2 × 1.5 Note that it would have been sufficient to specify m/l, the mass per EXAMPLE particles unit length of the wire. The earth’s 4 × 10–5 T and we have ignored it. magnetic field is approximately 4.1 moving demonstration: in a Example 4.2 If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis (Fig. 4.4), which way would the Lorentz force be for (a) an electron (negative magnetic charge), (b) a proton (positive charge). field. http://www.phys.hawaii.edu/~teb/optics/java/partmagn/index.html FIGURE 4.4 Solution The velocity v of particle is along the x-axis, while B, the magnetic field is along the y-axis, so v × B is along the z-axis (screw rule or right-hand thumb rule). So, (a) for electron it will be along –z EXAMPLE axis. (b) for a positive charge (proton) the force is along +z axis. 4.2 111 Reprint 2025-26 Physics 4.3 MOTION IN A MAGNETIC FIELD We will now consider, in greater detail, the motion of a charge moving in a magnetic field. We have learnt in Mechanics (see Class XI book, Chapter 5) that a force on a particle does work if the force has a component along (or opposed to) the direction of motion of the particle. In the case of motion of a charge in a magnetic field, the magnetic force is perpendicular to the velocity of the particle. So no work is done and no change in the magnitude of the velocity is produced (though the direction of momentum may be changed). [Notice that this is unlike the force due to an electric field, qE, which can have a component parallel (or antiparallel) to motion and thus can transfer energy in addition to momentum.] We shall consider motion of a charged particle in a uniform magnetic field. First consider the case of v perpendicular to B. The perpendicular force, q v × B, acts as a centripetal force and produces a circular motion perpendicular to the magnetic field. The particle will describe a circle if v and B are perpendicular to each other (Fig. 4.5). If velocity has a component along B, this component remains unchanged as the motion along the magnetic field will not be affected by the magnetic field. The motion in a plane perpendicular to B is as before a circular one, thereby producing a helical motion (Fig. 4.6). You have already learnt in earlier classes (See Class XI, Chapter 3) that if r is the radius of the circular path of a particle, then a force of m v2 / r, acts perpendicular to the path towards the centre of the circle, and is called the centripetal force. If the FIGURE 4.5 Circular motion velocity v is perpendicular to the magnetic field B, the magnetic force is perpendicular to both v and B and acts like a centripetal force. It has a magnitude q v B. Equating the two expressions for centripetal force, m v 2/r = q v B, which gives r = m v / qB (4.5) for the radius of the circle described by the charged particle. The larger the momentum, the larger is the radius and bigger the circle described. If w is the angular frequency, then v = w r. So, w = 2p n = q B/ m [4.6(a)] which is independent of the velocity or energy . Here n is the frequency of rotation. The independence of n from energy has important application in the design of a cyclotron. The time taken for one revolution is T= 2p/ w º 1/n. If there is a component of the velocity FIGURE 4.6 Helical motion parallel to the magnetic field (denoted by v||), 112 it will make the particle move along the field and the path of the Reprint 2025-26 Moving Charges and Magnetism particle would be a helical one (Fig. 4.6). The distance moved along the magnetic field in one rotation is called pitch p. Using Eq. [4.6 (a)], we have p = v||T = 2pm v|| / q B [4.6(b)] The radius of the circular component of motion is called the radius of the helix. Example 4.3 What is the radius of the path of an electron (mass 9 × 10-31 kg and charge 1.6 × 10–19 C) moving at a speed of 3 ×107 m/s in a magnetic field of 6 × 10–4 T perpendicular to it? What is its frequency? Calculate its energy in keV. ( 1 eV = 1.6 × 10–19 J). Solution Using Eq. (4.5) we find r = m v / (qB) = 9 ×10–31 kg × 3 × 107 m s–1 / ( 1.6 × 10–19 C × 6 × 10–4 T) = 28 × 10–2 m = 28 cm n = v / (2 pr) = 17×106 s–1 = 17×106 Hz =17 MHz. 2 EXAMPLE E = (½ )mv = (½ ) 9 × 10–31 kg × 9 × 1014 m2/s2 = 40.5 ×10–17 J ≈ 4×10–16 J = 2.5 keV. 4.3
11.2 — Electron Emission
Physics Class 12 · Chapter 11
11.2 ELECTRON EMISSION We know that metals have free electrons (negatively charged particles) that are responsible for their conductivity. However, the free electrons cannot normally escape out of the metal surface. If an electron attempts to come out of the metal, the metal surface acquires a positive charge and pulls the electron back to the metal. The free electron is thus held inside the metal surface by the attractive forces of the ions. Consequently, the electron can come out of the metal surface only if it has got sufficient energy to overcome the attractive pull. A certain minimum amount of energy is required to be given to an electron to pull it out from the surface of the metal. This minimum energy required by an electron to escape from the metal surface is called the work function of the metal. It is generally denoted by f0 and measured in eV (electron volt). One electron volt is the energy gained by an electron when it has been accelerated by a potential difference of 1 volt, so that 1 eV = 1.602 ×10–19 J. This unit of energy is commonly used in atomic and nuclear physics. The work function (f0) depends on the properties of the metal and the nature of its surface. The minimum energy required for the electron emission from the metal surface can be supplied to the free electrons by any one of the following physical processes: (i) Thermionic emission: By suitably heating, sufficient thermal energy can be imparted to the free electrons to enable them to come out of the 275 metal. Reprint 2025-26 Physics (ii) Field emission: By applying a very strong electric field (of the order of 108 V m–1) to a metal, electrons can be pulled out of the metal, as in a spark plug. (iii) Photoelectric emission: When light of suitable frequency illuminates a metal surface, electrons are emitted from the metal surface. These photo(light)-generated electrons are called photoelectrons. 11.3 PHOTOELECTRIC EFFECT 11.3.1 Hertz’s observations The phenomenon of photoelectric emission was discovered in 1887 by Heinrich Hertz (1857-1894), during his electromagnetic wave experiments. In his experimental investigation on the production of electromagnetic waves by means of a spark discharge, Hertz observed that high voltage sparks across the detector loop were enhanced when the emitter plate was illuminated by ultraviolet light from an arc lamp. Light shining on the metal surface somehow facilitated the escape of free, charged particles which we now know as electrons. When light falls on a metal surface, some electrons near the surface absorb enough energy from the incident radiation to overcome the attraction of the positive ions in the material of the surface. After gaining sufficient energy from the incident light, the electrons escape from the surface of the metal into the surrounding space. 11.3.2 Hallwachs’ and Lenard’s observations Wilhelm Hallwachs and Philipp Lenard investigated the phenomenon of photoelectric emission in detail during 1886-1902. Lenard (1862-1947) observed that when ultraviolet radiations were allowed to fall on the emitter plate of an evacuated glass tube enclosing two electrodes (metal plates), current flows in the circuit (Fig. 11.1). As soon as the ultraviolet radiations were stopped, the current flow also stopped. These observations indicate that when ultraviolet radiations fall on the emitter plate C, electrons are ejected from it which are attracted towards the positive, collector plate A by the electric field. The electrons flow through the evacuated glass tube, resulting in the current flow. Thus, light falling on the surface of the emitter causes current in the external circuit. Hallwachs and Lenard studied how this photo current varied with collector plate potential, and with frequency and intensity of incident light. Hallwachs, in 1888, undertook the study further and connected a negatively charged zinc plate to an electroscope. He observed that the zinc plate lost its charge when it was illuminated by ultraviolet light. Further, the uncharged zinc plate became positively charged when it was irradiated by ultraviolet light. Positive charge on a positively charged zinc plate was found to be further enhanced when it was illuminated by ultraviolet light. From these observations he concluded that negatively charged particles were emitted from the zinc plate under the action of ultraviolet light. After the discovery of the electron in 1897, it became evident that the incident light causes electrons to be emitted from the emitter plate. Due276 Reprint 2025-26 Dual Nature of Radiation and Matter to negative charge, the emitted electrons are pushed towards the collector plate by the electric field. Hallwachs and Lenard also observed that when ultraviolet light fell on the emitter plate, no electrons were emitted at all when the frequency of the incident light was smaller than a certain minimum value, called the threshold frequency. This minimum frequency depends on the nature of the material of the emitter plate. It was found that certain metals like zinc, cadmium, magnesium, etc., responded only to ultraviolet light, having short wavelength, to cause electron emission from the surface. However, some alkali metals such as lithium, sodium, potassium, caesium and rubidium were sensitive even to visible light. All these photosensitive substances emit electrons when they are illuminated by light. After the discovery of electrons, these electrons were termed as photoelectrons. The phenomenon is called photoelectric effect.
11.3 — The Photoelectric Cut-Off Voltage In A Certain Experiment Is 1.5 V.
Physics Class 12 · Chapter 11
11.3 The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
📋 Question Details
- Chapter
- Current Electricity
- Topic
- Electric Current
- Year
- 2024
- Shift
- 06 Apr Shift 2
- Q Number
- Q11
- Type
- MCQ
- NCERT Ref
- Class 12 Physics Ch 3: Current Electricity
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