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Phy 204: Electricity and Magnetism Name: Fall Semester 2014 Roll No: Final Exam: Part A Maximum Marks: 30 Total Time: 1 Hour 15 Minutes Instructions. Write your name and roll number in the space specified above. Attempt all questions. This exam comprises of two parts, A and B and will last for 3 hours. The recommended time for part A is 1 hour. Part A has 10 Multiple Choice Questions. The most appropriate answer is to be circled on the question paper. Part A is to be filled on the question paper and returned within 1 hour 15 minutes. You can use the blue answer books for rough work relevant to Part A. Please turn off mobile phones. List of Physical Constants Bohr Magneton, µB = 9.27 x 10−24 J K−1 Boltzmann Constant, kB = 1.38 x 10−23 J K−1 Vacuum Magnetic Permeability, µ0 = 4π x 10−7 T m A−1 Elementary Charge, e = 1.6 x 10−19 C Vacuum Permittivity, εo = 8.85 x 10−12 C2 N m−2 Useful Formulae ⃗ = µ0 q⃗v x ⃗r B 4π r2 ⃗ ⃗j = σ E (relation between current density and electric field); j = nqvd σ= (Biot-Savart Law); (current density depends on the charge density and drift velocity); nq 2 τ m (conductivity depends on average scattering time); ⃗ ⃗vd = µE (definition of mobility); ⃗ = µ0 nI |B| (magnetic field inside an infinite solenoid); ⃗ ⃗ ⃗= ExB S µ0 (Poynting Vector as a cross-product of electric and magnetic fields); Maxwell’s Equations: ∫ ⃗ dA ⃗ = Qenc E. ε0 (Gauss’s Law for Electric Charges); Closed Area ∫ ⃗ dA ⃗= 0 B. (Gauss’s Law for Magnetism); Closed Area ∫ ⃗ dA ⃗ B. (Faraday’s Law); Open Area Closed P ath { ∫ ⃗ d⃗l = µ0 B. Closed P ath ∫ ⃗ d⃗l = − d E. dt d Ienc + ε0 dt ∫ } ⃗ dA ⃗ E. Open Area (Maxwell-Ampere’s Law); MCQs Part A: 1. A wire is fashioned into a right angled triangle with two sides of equal length l. Initially, the triangle, which is at rest, is fully immersed in a uniform magnetic field B, normal to the plane of the wire. Its vertical side coincides with the edge of the magnetic field. At t = 0, the loop starts moving out of the field with constant speed v. This is shown in the diagram. What can we say about the induced emf ε(t) at time t? Region of non-zero B B = 0 here t=0 l l v (a) |ε(t)| = Bv(l − vt) for all t > 0 l l , ε(t) = 0 for t > v v l l (c) |ε(t)| = Bv(l − vt) for 0 < t < , ε(t) = 0 for t > v v (b) |ε(t)| = Bvl for 0 < t < (d) |ε(t)| = Bvl for t > 0 (e) |ε(t)| = Bvl2 for t > 0 t t>0 2. A coil of N turns is placed inside an electromagnet. The coil has radius r and is connected to a resistor. The total resistance inside the circuit is R. The magnetic field is confined to the area of the coil (as shown). The electromagnet is switched off and the field decays to zero in time τ . How much total charge Q flows through the resistor? N turns r Field B is only inside the loop R (a) An emf and a current will be induced but Q = 0. N Bπr2 τ R N Bτ (c) Q = R N Bπr2 (d) Q = Rτ N Bπr2 (e) Q = . R (b) Q = 3. An infinite solenoid of diameter D carries a linearly increasing current I(t) = kt, where k > 0. Point P is inside the solenoid at a distance r ≪ D from the axis of the solenoid. Point Q is outside the solenoid. Which of the following graphs show the magnitude of the electric field at points P and Q? Assume the field is uniform inside the solenoid. Q Cross-section of the infinite solenoid P r D EP EQ EQ is constant and non-zero EP is constant and non-zero 0 t 0 t (a) EP EQ EP is constant and non-zero 0 (b) t EP EQ= 0 always 0 EQ EQ= 0 always EP = 0 always 0 (c) t t 0 t EP EQ EP increases quadratically 0 0 t (d) EP t EQ EQ= 0 always 0 0 t (e) t 4. An infinite solenoid carries a time-varying current I(t) = I0 cos(ωt). A flexible conducting ring with smaller radius (as shown) is placed inside and coaxially with the solenoid. What will happen to the ring? Ring I(t) x x x x x shrink stretch x x x x x Current coming out Current going in (a) The ring will continuously stretch and increase its radius. (b) The ring will alternately stretch and shrink. (c) The ring will continuously shrink and decrease its radius. (d) The ring will twist because it will experience a torque. (e) The ring will start rotating about the axis of the solenoid. 5. A constant current I flowing through a conducting wire terminates on a point charge Q, depositing more and more charge as time passes by. What is the rate of change of electric flux dΦe through a closed surface (represented by the dashed line in the figure) enclosing the point dt charge? Q I µ0 I 4πε0 r2 I (b) 4πε0 r2 µ0 ε0 I (c) 4πr2 I (d) ε0 (a) (e) I(4πr2 ) r 6. A positive charge of +Q is uniformly distributed over the surface of a hollow, insulating sphere of radius L. A point charge +Q is located at a distance 4L from the center. Point A is located at a distance L from the sphere’s surface, on the axis joining the point charge and the center L of the sphere. Point B is located at a distance from the center of the sphere. What can be 2 said about the electric fields at A and B? + + charge +Q 2L A + L + + L + + +Q distributed + + uniformly + + + L/2 + B + + + (a) EA = 0, EB = 0 (b) EA = 0, EB ̸= 0 (c) EA ̸= 0, EB ̸= 0 (d) EA ̸= 0, EB = 0 (e) |EA | < |EB | 7. In experiment A, three point charges, each of magnitude +Q are placed symmetrically on the perimeter of a circle. In experiment B, a charge 3Q is uniformly distributed over the same circle. Which of the following statements is true about the magnitudes of the electric fields at different points on the dashed circle, concentric with the bold circle? Charge 3Q +Q + + + + + + +Q +Q + + + + + + B A (a) Magnitude of the electric field varies from point to point on the dashed circles both in experiment A and B. (b) Magnitude of the electric field is the same at all points on the dashed circle in A and at all points on the dashed circle in B, but it differs between A and B. (c) Magnitude of the electric field is the same at all points on the dashed circle in A and at all points on the dashed circle in B, and it is the same in A and B. (d) Magnitude of the electric field is the same everywhere on the dashed circle only in experiment A. (e) Magnitude of the electric field is the same everywhere on the dashed circle only in experiment B. 8. A high frequency alternating current flows through a uniform wire. What is an approximate pattern of the electric field inside the wire at the particular time t0 , shown in the accompanying diagram? The arrows represent electric field vectors. Dots represent zero electric field. [HINT: Skin Effect] I (t) t0 time (a) (b) (c) (d) (e) 9. A switch inside an RC circuit is closed at time t = 0. The ideal capacitor builds up charge with ⃗ in the region between the capacitor time. Which of the following shows the Poynting Vector S plates while the capacitor charges? ε a b a S b (a) a S b (b) a X X X S b (c) a S b (d) ⃗ = 0) (e) There is no energy flow into or out of the capacitor (S 10. A metallic conductor has a hot and a cold end. The resistivity depends on temperature. How ⃗ A and E ⃗ B at points A and B compare? do the electron mobilities µ ⃗ A and µ ⃗ B and electric fields E Hot Cold A Heating by laser ⃗ A | > |E ⃗B| (a) |⃗ µA | < |⃗ µB |, |E ⃗ A | = |E ⃗B| (b) |⃗ µA | = |⃗ µB |, |E ⃗ A | > |E ⃗B| (c) |⃗ µA | > |⃗ µB |, |E ⃗ A | < |E ⃗B| (d) |⃗ µA | > |⃗ µB |, |E ⃗ A | = |E ⃗B| (e) |⃗ µA | > |⃗ µB |, |E B Phy 204: Electricity and Magnetism Name: Fall Semester 2014 Roll No: Final Exam: Part B Solution Maximum Marks: 40 Total Time: 1 hour 45 minutes Instructions. Write your name and roll number in the space specified above. Attempt all questions. This exam comprises of two parts, A and B. There are 6 comprehensive questions in Part B. Use the blue answer books provided to answer Part B questions as completely as possible. Please write your name and roll number on ALL answer books. List of Physical Constants Bohr Magneton, µB = 9.27 x 10−24 J K−1 Boltzmann Constant, kB = 1.38 x 10−23 J K−1 Vacuum Magnetic Permeability, µ0 = 4π x 10−7 T m A−1 Elementary Charge, e = 1.6 x 10−19 C Vacuum Permittivity, εo = 8.85 x 10−12 C2 N m−2 Useful Formulae ⃗ = µ0 q⃗v x ⃗r B 4π r2 ⃗ ⃗j = σ E (relation between current density and electric field); j = nqvd σ= (Biot-Savart Law); (current density depends on the charge density and drift velocity); nq 2 τ m (conductivity depends on average scattering time); ⃗ ⃗vd = µE (definition of mobility); ⃗ = µ0 nI |B| (magnetic field inside an infinite solenoid); ⃗ ⃗ ⃗= ExB S µ0 (Poynting Vector as a cross-product of electric and magnetic fields); Maxwell’s Equations: ∫ ⃗ dA ⃗ = Qenc E. ε0 (Gauss’s Law for Electric Charges); Closed Area ∫ ⃗ dA ⃗= 0 B. (Gauss’s Law for Magnetism); Closed Area ∫ ⃗ dA ⃗ B. (Faraday’s Law); Open Area Closed P ath { ∫ ⃗ d⃗l = µ0 B. Closed P ath ∫ ⃗ d⃗l = − d E. dt d Ienc + ε0 dt ∫ } ⃗ dA ⃗ E. Open Area (Maxwell-Ampere’s Law); Part B: Comprehensive Questions ⃗ pointing downwards exists only in a thin columnar region of radius 1. A uniform magnetic field B b. A positively charged ring of uniform charge density λ is coaxial with this column. The field dB dB is switched off slowly at a rate . Note that < 0. dt dt B b + + + + + + + + λ = Q/2πr r + + + + + + + (a) What happens to the ring of charge? (i) It will start rotating in the clockwise direction (as seen from above). (ii) It will start rotating in the anticlockwise direction (as seen from above). [Correct Answer= 2 marks, Incorrect = -1 mark, Blank = 0 marks.] ⃗ has dropped to 0. First (b) Find the total angular momentum L imparted to the ring when B find the torque τ exerted on the charged ring due to the changing magnetic field. [7 marks] ∫ (HINT: L = τ dt where L is the angular momentum and τ is the torque.) Answer Step 1: A time-varying magnetic field creates a non-conservative electric field (Faraday’s Law). We first find the electric field generated at the distance r. The right-hand curl rule suggests that this electric field will be circular and in the clockwise direction (as seen from above). ∫ ∫ ⃗ d⃗l = − d E. dt Closed P ath ⃗ dA ⃗ B. Open Area dB E.2πr = − π b2 dt dB πb2 E = − . dt 2πr Step 2: We can now find the electric force acting on the ring of charge. Torque exerted on the ring τ (t) is then calculated from the electric force: dF = Edq dτ τ = |d⃗τ | = |⃗r x dF⃗ | = rdF = Erdq ∫ = Er dq = ErQ τ = − dB πb2 rλ(2πr) dt 2πr dB = − (πb2 ) rλ. dt Step 3: Calculate the total angular momentum L imparted to the ring using the formula provided: ∫ L = τ dt ∫0 −(πb2 ) rλdB = B = (πb2 ) rλB. 2. We have two infinite, thin sheets, each carrying a uniform current of sheet density κ (where κ is in Amperes per meter). The currents are in opposite directions, as shown below. Copy the diagram and mark the direction of the magnetic field due to this current distribution on your diagram. Plot the magnitude of this magnetic field as a function of x. Calculate and identify the maximum and minimum field on your plot. Explain your reasoning. [7 marks] Sheet A Current flowing out x is negative x= 0 Sheet B x x x x x Current x flowing in x x x x x x= d x Answer We can calculate the magnetic field in each region using Ampere’s Law. For example, consider the Amperian loop shown below for Sheet A. The loop is of width w and length l and is traversed in the direction shown, starting from point P . Assuming that the magnetic field produced by Sheet A is B0 , we can calculate B0 as follows: Sheet A w B0 Current flowing out l B 0 P ∫ ⃗ ⃗l = µ0 Ienc B.d B0 (l) + 0 + B0 (l) + 0 = µ0 (κl) 2B0 (l) = µ0 (κl) B0 = µ0 κ . 2 Using the right hand curl rule, we can verify that the direction of the magnetic fields produced by both sheets (labeled as sheet A and B) are as shown below. Note that the magnetic fields are independent of the distance from the sheet. The magnetic fields cancel out in regions x < 0 and x > d and add in the region 0 < x < d due to superposition. Sheet A BA A B x BA x x x x Current x flowing in x x x x x Current flowing out x is negative Sheet B x x= d x= 0 From the Principle of Superposition: BT otal = B0 − B0 = 0 (x < 0) BT otal = B0 + B0 = µ0 κ (0 < x < d) BT otal = B0 − B0 = 0 (x > d). This is shown in the graph below. The maximum field is µ0 κ and the minimum field is zero. BTotal µ0κ 0 0 d x 3. A cylindrical conductor carries a uniform current of density ⃗j. Find the ratio of the magnitudes of the Poynting Vectors at the surface and half-way between the surface and the central axis of the cylinder. [5 marks] j R/2 R Answer ⃗ = S S = ⃗ xB ⃗ E µ0 EB . µ0 Using Ampere’s Law, calculate the magnetic field at the surface (B1 ): B1 (2πR) = µ0 j(πR2 ) B1 = µ0 j(πR2 ) . 2πR Similarly, calculate the magnetic field half-way from the surface (B2 ): B2 (2π R R2 ) = µ0 j(π ) 2 4 µ0 j(πR2 ) B2 = . 4πR ⃗ is the same throughout the conductor (since ⃗j is constant and ⃗j = σ E). ⃗ Therefore, S1 = E S2 4 B1 = = 2. B2 2 4. You will not receive any credit for using the Biot-Savart Law in this question. A U-shaped wire is made up of two parallel straight wire segments of semi-infinite length placed a distance 2R apart and a semi-circular segment of radius R connecting the straight wires, as shown in the diagram below. A constant current I passes through this wire. Using Ampere’s Law and the Principle of Superposition, find the magnitude and direction of the magnetic field ⃗ at point C, the center of the semi-circular segment of the wire. [6 marks] B You are given that the the magnetic field at the center of a circular loop of wire of radius R, µ0 I carrying current I, is . 2R I C R R Answer Consider the figures drawn below, showing A and B which are identical with respect to the field they produce at C and C ′ . I C C' = B A I I C R A+B = P I Let’s first find the field at C in the configuration P as shown above. Later we will halve this result. The configuration P is a sum of two infinitely long wires and a circular loop of wire, all three carrying a current I. Using Ampere’s Law it can be found that the magnetic field µ0 I at distance R due to an infinite wire carrying current I is . Using the Right Hand Grip 2πR rule, it is found that all three fields are in the same direction, hence are added (Principle of Superposition). P = = = µ0 I µ0 I µ0 I + + 2πR 2πR 2R µ0 I µ0 I + πR ( 2R ) µ0 I 1 1 + . R π 2 Field in configuration A is half of the field in configuration P (Principle of Superposition), ( ) µ0 I 1 1 + . hence the required field is 2R π 2 5. A conducting sphere of radius R initially has a uniform surface-charge density σ0 . Beginning ( ) t at t = 0, this charge is drained off over a period t0 such that σ = σ0 1 − . t0 R σ Ic (a) Find the magnitude of the displacement current Id just outside the surface of the sphere. [6 marks] (b) Calculate the current Ic carried off by the wire. Compare the displacement current Id to Ic . [4 marks] Answer Id R σ Ic We use Maxwell-Ampere’s Law in this question. We choose a very small circular path encircling the wire carrying the conduction current. This path is shown in red in the diagram above. For part (a) we consider the spherical surface just outside the conducting sphere as the open surface corresponding to this path. Note that Ic through this surface (shown as a dashed circle in the diagram above) is zero. In part (b) we should be able to verify that Id = Ic , since the only current piercing the flat circular surface enclosed by the red path is Ic . (a) Id d = ε0 dt ∫ ⃗ dA ⃗ E. Open Area = ε0 dE (4πR2 ). dt dE . Using Gauss’s Law for Electric Charges we find the Electric Field outside dt the sphere in terms of the electric charge density: We need to find ∫ ⃗ dA ⃗ = E. Qenc ε0 Closed Area E(4πR2 ) = Q ε0 = (4πR2 ) ⇒E = σ ε0 σ . ε0 This gives us dE dt Id = − 1 dσ ε0 dt ( ) 1 1 (σ0 ) − = ε0 t0 σ0 . = − ε 0 t0 = (4πR2 )σ0 (4πR2 )σ0 and |Id | = . t0 t0 (b) dQ dt ( ) d t 2 = (4πR )σ0 1 − dt t0 2 (4πR )σ0 = − . t0 Ic = Therefore, |Ic | = |Id |. Note that this was expected from the Maxwell-Ampere’s Law. Both currents equal the path integral of the magnetic field around the small circular red path, therefore must be equal. ⃗ = B0 xî, that is, a uni-directionally 6. Using a Maxwell’s Equation show that a magnetic field B varying magnetic field (as shown below) cannot exist. [3 marks] B y x z Answer B1 B2 B y x z Using Gauss’s Law for Magnetism it can be shown that a uni-directionally varying magnetic field cannot exist. We draw a Gaussian Surface as shown above. Note that since there are no y or z components of the magnetic field, the flux through the curved surface area of the cylinder is zero. Taking A to be the area of the end caps we find: ∫ ⃗ dA ⃗ = B. 0 Closed Area B2 A − B1 A = 0 ⇒ B1 = B2 . Hence B1 cannot be different from B2 . Therefore a uni-directionally varying magnetic field, which changes with x, cannot exist because it violates Gauss’s Law for magnetism.