Magnetic effect of_current


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Magnetic effect of_current

  1. 1. Dual Nature of Matter and Radiation HOTS Questions 1 Mark Questions and Answers 1. An uncharged zinc plate becomes positively charged when irradiated by ultraviolet radiations. What is the phenomenon due to? Answer The phenomenon of electron emission is called photoelectric effect. 2. In an experiment on photoelectric effect, the photoelectric current (I) and the anode, the following graphs were obtained between potential (V). Name the characteristic of the incident radiation that was kept constant in this experiment. I O V Answer Frequency of incident radiation was kept constant. 3. Are matter waves electromagnetic? Answer No, Matter waves are not electromagnetic as they are associated with neutral particles. They are probability waves which tell the probability of location of particle in a certain region of space.
  2. 2. 4. A student plots graph between de-Broglie wavelength and some power (n) of momentum. What is the value of n? λ pn Answer We have λ= h/p = h.p-1 → λ∞ p-1 n= -1 5. Represent graphically the variation of the de-Broglie wavelength with linear momentum of the particle. Answer We have λ = h/p λ∞1/pλ The graph between h and p is shown p 2 Marks Questions and Answers 1. An electron and an alpha particle have the same de-Broglie wavelength associated with them. How are their kinetic energies related to each other? Answer λelectron=λα de-Broglie wavelength associated with a particle of mass m and energy E is λ = h/√2mE h/√2meEe = h/√2maEa That is kinetic energy of electron and a particle are in inverse ratio of these masses.
  3. 3. 2. Electrons are emitted from the surface when green light is incident on it, but no electrons are ejected when yellow light is incident on it. Do you expect electrons to be ejected when surface is exposed to (i) red light (ii) blue light? Answer (i) The wavelength of red light is longer than threshold wavelength hence no electron will be emitted with red lights. (ii) The wavelength of blue light is smaller than threshold wavelength hence will be ejected. 3. Draw a graph showing the variation of stopping potential with frequency of incident radiation in relation to photoelectric effect. Deduce an expression for the slope of this graph using Einstein’s photoelectric equation. Answer From Einstein’s photoelectric equation Ek = hv – hv0 V0 eV0 = hv – hv0 Vo = (h/e).v –(h/e).v0v0 ν Clearly Vo – v graph is a straight line of form y = mx + c; the slope of graph is m = h/e 4. Two lines A and B shown in the graph represent the de-Broglie wavelength (λ) as a function of 1/√V (V is the accelerating potential) for two particles having the same charge. Which of the two represents the particle of smaller mass? B λ A 1/√V
  4. 4. Answer De-Broglie wavelengthλ = h/√2mqV or λ = h/√2mq . 1/√V The graph of λ versus 1/√V is a straight line of slope h/√2mq α 1/√m. the slope of line B is large. So particle B has smaller mass. 5. If the frequency of light falling on a metal is doubled, what will be the effect on photo current and the maximum kinetic energy of emitted photoelectrons? Answer The photocurrent does not depend on the frequency of incident radiation, hence the photocurrent remains unchanged. The maximum kinetic energy increases with increase of frequency, given by EK = hν – W If frequency is doubled, EK’ = 2 hν– W EK’ / EK= ( 2 hν – 2W +W) / (hν – W) = 2 + W / (hν – W) ˃ 2 i.e maximum kinetic energy will increase to slightly more than double value. 3 Mark Questions and Answers: 1. Radiations of frequency 1015 Hz are incident on two photosensitive surfaces A and B. following observations are recorded. Surface A: No Photoemission takes place. Surface B: Photoemission takes place but photoelectrons have zero energy. Explain the above observations on the basis of Einstein’s photoelectric equation. How will the observation with surface B change when wavelength of incident light is decreased? Answer Einstein’s Photoelectric equation is Hν = W + Ek→Ek= hν – W orEk = hν – hν0 Where W is work function of metal, ν is frequency of incident light and ν0 is threshold frequency.
  5. 5. Surface A: As no photoemission takes place; energy of incident photon is less than the work function. In other words, the frequency v = 1015 Hz of incident light is less than the threshold frequency (v0) Surface B: As phenomenon takes place with zero kinetic energy of photoelectrons (i.e, Ek = 0), then equation (1) gives W = hv or v0 = v. i.e., energy of incident photon is equal to work function. In other words, threshold frequency of metal is equal to frequency of incident photon i.e., v0 = 1015 Hz. When wavelength of incident light is decreased, the energy of incident photon becomes more than the work function, so photoelectrons emitted will have finite kinetic energy given by Ek = (hc/λ )– W. 2. In a photoelectric experiment, an incident radiation of wavelength λ emits photo electrons with maximum kinetic energy E. Show that to get the electrons with energy 2E; the wavelength of the incident radiations should be hcλ/(Eλ + hc)? Answer We have E = hv – φ0 = (hc/λ ) - φ0 For electron with energy 2E, let the wavelength be λ1 Then 2E = (hc/λ1) - φ0 or(2hc/λ)- 2φ0 = (hc/λ1) - φ0 hc/λ1 = (2hc/λ) - φ0 = 2hc/λ + E – hc/λ = hc/λ + E = (hc + Eλ)/λ λ1= hcλ/(Eλ + hc) 3. Assuming that 1 % of the photon in a 39.8 Wm-2 intensity beam of wavelength 5000 Å cause emission of photoelectrons, calculate the resulting photoelectric current ? Answer Let the total number of photons in the beam/sec be n Then nhv = 39.8 Or nhc/λ = 39.8 N = 39.8 x λ/hc = 39.8 x 5000 x 1010 / (6.63 x 10-34 x 3 x 108) = (199/19.89) x 1019 = 1020 No of photoelectrons emitted/sec = no of photons absorbed = 1 % of 1020 = 1018 Current = 1018e = 1018 x 1.6 x 10-19 = 0.16 A = 160mA.
  6. 6. 4. The human eye needs a minimum intensity of 10-10 Wm-2 to produce sensation of sight. If the area of the pupil is 10-6 m2 and the wavelength of incident light is 5600 Å; what should be the minimum number of photons entering the pupil / sec to excite the sense of sight? Answer Minimum intensity detectable = 10-10 W/m2 Minimum energy required on pupil = 10-10 W/m2 x 10-6 m2 = 10-16 joule/sec Let the number of photons entering the eye be n/sec Then n.hv = 10-16 n = 10-16/(6.63 x 10-34 x c/λ) = 10-16 x 5600 x 10-10/(6.63 x 10-34 x 3 x 108) = 5600/19.89 = 283 5. An electron moving with velocity v and a photon have same de-Broglie wavelength. What will be the ratio of their energy values? Answer We have λe =λph h/Pe = h/Pph→Pe = Pph ormv = Eph/c (1) Now 1/2mv2 = Ee→ m = 2Ee/v2 From (1) 2Ee/v = Eph/c Hence Ee/Eph = v/2c. Prepared by Mrs. V.Balasaraswathy PGT (Physics) KV-NA
  7. 7. Magnetic effect of current HOTS 1. Figure shows two current carrying curves 1 and 2. Find the magnitudes and directions of the magnetic field at points p, Q and R. Ans: According to right hand gap rule, B1 of wire 1 at point p will point normally outward while B2 of wire 2 will point normally inward. BP = B1- B2 =𝜇0I1/2𝜋r1-𝜇0I2/2𝜋r2 = 4𝜋𝑋10−7 2𝜋 (20/0.10 -30/0.3) = 2 X 10-5 T pointing normally outward. 2. Two infinitely long insulated wires are kept perpendicular to each other. They carry currents I1= 2A and I2=1.5 A. (i) Find the magnitude and direction of the magnetic field at p. (ii) If the wires , what would reserved in one of the wires, what would be the magnitude of the field B?
  8. 8. Ans: B1= 𝜇0I1/2𝜋r1= 4𝜋𝑋10−7𝑋2 2𝜋𝑋4𝑋10−2 =10-5 T. Normally into the plane of the paper. B2= 𝜇0I2/2𝜋r2= 4𝜋𝑋10−7𝑋1.5 2𝜋𝑋3𝑋10−2 =10-5 T normally into the plane of paper B= B1+B2= 2 X 10-5 T. 3. Two co-axial circular loops L 1 and L2 of radii 3cm and 4cm are placed as shown. What should be the magnetic and direction of the current in the loop L2 so that the net magnetic field at the point o be zero? For the net magnetic field at the point ‘o’ to be zero, the direction of current in loop L2 should be opposite to that in loop L1. Magnitude of magnetic = Magnitude of magnetic field due to current I1 in L1 field due to current I2 in L2 𝜇0 𝐼1(0.03)2 2[(0.03)2 + (0.04)2] 3 2 = 𝜇0 𝐼2(0.04)2 2[(0.03)2 + (0.04)2] 3 2
  9. 9. I2 = 9/16 X 1 = 0.56 A 4. A straight wire carrying a current of 12A is bent into a semicircular are of radius 2 cm as shown in figure. What is the direction and magnitude of 𝐵⃗ at the centre of the arc? Would your answer change if the wire were bent into a semicircular arc of the same radius bent in the opposite way as shown in the figure? (i) B at the centre of the arc is B = 𝜇0 𝐼 4𝑟 𝐵 = 4𝜋𝑋10−7 𝑋12 4 𝑋 0.02 = 1.9 𝑋 10−4 𝑇 According to right hand rule, the direction of the field is normally into the plane of paper. ii) The magnetic field will be of the same magnitude B = 1.9 X 10-4 T The direction of the field is normally out of the plane of paper. 5. A beam of electron projected along +x –axis experiences a force due to a magnetic field along the +y- axis. What is the direction of the magnetic field. V along X-axis- 𝑖.̂ B along y- axis-𝑗.̂
  10. 10. F = q ( V X B). V X B is along 𝑖̂𝑋 𝑗̂ = 𝑘.̂ i.e. +Z – axis. 6. A long straight wire AB carries a current I. A proton p travels with a speed V1 parallel to the wire, at a distance d from A in a direction opposite to the current as shown in the figure. What is the force experienced by the proton and what is its direction? Ans: B = 𝜇0I/2𝜋d. F = qvB= q X V X 𝜇0I/2𝜋d. F = 𝜇0Iqv/2𝜋d Direction towards right. MAGNETIC EFFECT OF CURRENT VALUE BASED QUESTIONS
  11. 11. 1.MrNarasimhan a 65 year old person often complained of neck pain. One day his grandson Avinash, suggested that magnetic therapy is very effective in reducing such pains. He said that the permanent magnet/electromagnet , used in the device will help to produce Joule’s heating effects in the blood stream, which helps the blood flow better.He immediately contacted his friend in Chennai, who was running Magnetic Therapy Clinic. Mr Narasimhan who felt better. (a).What two values did Avinash exhibit towards his grandfather? Mention any two (b). What is the SI unit of magnetic induction and define it? Ans. (a)Responsiblebehaviour, concern and awareness (b) Tesla (defn) 2. MsUdaya joined a PG course in Nanotechnology lab in IIT Chennai. The first day, when she went to the lab, she met Mr. Antonio, the lab assistant.He greeted her and advised her not to touch the wires which were suspended from the roof at every part of the lab as they were from high voltage lines. He also told her not to bring any of the two wires closer to each other during any experimental applications. He helped her in understanding about the precautions that has to be taken in the lab.
  12. 12. (a) What value did Mr. Antonio exhibit towards Ms. Udaya? Mention any two (b)Why two high voltage power transmission lines should not be close to each other? (a) Responsible behaviour, sensitivity, concern for others and alerting the People 3. In the birthday party of Bharat, a class 7student, his parents gave big slinkys to all his friends as return gifts. The next day, during the physics classMr Mohan, the teacher explained them about the production of magnetic fields using current carrying coil and also said that they can make magnets, using such coils by passing high currents through them. That night Sumanth, a friend of Bharat, asked his father about the coils, and their shape. His father asked him to bring the slinky, that his friend gave and explained the uses of toroid and solenoid. (a)What value did Sumanth’s father exhibit towards his son? (b).What is the difference in the fields produced by the solenoid and Toroid? Ans(a)Responsibility, makes his child to understand the concepts and to generate interest in the subjects b)The magnetic field lines in a toroid is concentric circles whereas in solenoid it is straight within the turns.
  13. 13. 4 .MsLavanya a house wife aged 42 years complained of stomach ache one day. Her husband MrSrinivas took her to a nearby hospital. The doctor observed her and found something wrong near her liver and suspected malignancy. There after checking her MRI scan, a team of doctors advised her to go through Carbon radio therapy which is very safe. They said using cyclotron, high speed ions can be generated that directly attach the cancerous tissues and destroy them. (a). What values did Mrsrinivas and the doctor have exhibited? Mention any two. (b).What are the role played by Electric field and magnetic fielding Cyclotron? Ans.(a) concern for others, helpfulness, presence of mind, responsible citizen (b) The charged particles are accelerated by the electric field and magnetic field provides the necessary centripetal acceleration.