Notes and Solved Problems for Common Exam 3 (Does not include Induction)

Notes and Solved Problems for Common Exam 3 (Does not include Induction) 8. MULTI LOOP CIRCUITS Key concepts: Multi loop circuits of batteries and resistors: loops, branches and junctions should be distinguished. A general problem is to find currents insides every branch if the resistances and EMF s of the batteries are known. Three rules apply: Branch rule: inside every branch the current is the same. The current may be different in different branches. Junction rule: the sum of all currents coming into the junction is equal to the sum of all currents leaving the junction. Loop rule: the sum of all potential differences in any complete walk through any closed loop of the circuit is equal to zero. The strategy for solving multi-loop circuit problems is the following: 1. Find all branches and enumerate currents. 2. Find all junctions and establish the relationships between the currents 3. Apply Kirchoff loop rules to the loops in the circuit. The total number of equations to be written should be equal to the total number of currents, which are unknown. Typical problems related to multi loop circuits: Problem 5. What is the current inside the resistor R 1? What is the power released in the resistor R 1? A. 1A, 27 W B. 2A, 13 W C. 3 A, 27 W D. 4 A, 12 W E. 5 A, 5 W Solution. While this is a multi loop circuit, the answer to that

particular question can be found very easily. We see that the first loop contains the battery E 1 and only this resistor. Let us we write the Kirchoff loop rule for this particular loop. The current is assumed to be directed down inside R 1, therefore we walk through the loop clockwise: E 1 -i 1 *R 1 =0 From which we figure out that i 1 =E 1 /R 1 =9/3=3 A. The power released in the resistor (energy rate dissipated in the resistor) is P=i 1 *V=i 1 *i 1 *R=i 1 2 *R=3*3*3=27 W. This trick works when you can find such a loop in the multi loop circuit which contains only one resistor and (possibly many) batteries. Then you can apply the Kirchoof loop rule to this particular loop and find the current inside the resistor immediately! Answer C. Problem 6. Order branches of bulbs by brightness, dimmest first: A. I, II, III B. II, III, I C. III, I, II D. III, II, I E. II, I, III Solution. The brightness of each branch is proportional to the total power released in the bulbs of the branch. Assume that the resistance of the bulbs is the same and it is equal to R. Then the total power of branch I is: P 1 = i 1 *V 1 =( i 1 ) 2 *R, the total power of branch II is: P 2 = ( i 2 ) 2 *R + ( i 2 ) 2 *R = 2*( i 2 ) 2 *R, and the total power of branch III is: P 3 = ( i 3 ) 2 *R + ( i 3 ) 2 *R + ( i 3 ) 2 *R =3*( i 3 ) 2 *R. Note that the power is expressed via the current and resistance here. What about the currents i 1, i 2, i 3? We can find currents easily by applying the Kirchoff loop equations to the loops which accounts for the battery and the particular branch. So, that for the branch I and the battery the equation is E- i 1 *R=0 For the branch II and the battery, the equation is E- i 2 *R- i 2 *R =0 For the branch III and the battery the equation is E- i 3 *R- i 3 *R- i 3 *R =0 Therefore i 1 =E/R, i 2 =E/(2R), i 3 =E/(3R)

The resulting powers are P 1 = ( i 1 ) 2 *R=E 2 /R P 2 = 2*( i 2 ) 2 *R=2 E 2 /(4R)= E 2 /(2R) P 3 =3*( i 3 ) 2 *R=3E 2 /(9R)= E 2 /(3R) We finally see that the brightest branch is I (has largest power), then II, and finally III. Therefore, the dimmest is III, then II, then I. Answer D. Problem 7. In the previous problem (see figure to the problem 6), if E=12 V and R= 2Ω, find the power released in each bulb? A. 8W for branch I, 18W for branch II, 72W for branch III B. 18W for branch I, 8W for branch II, 72W for branch III C. 8W for branch I, 72W for branch II, 18W for branch III D. 72W for branch I, 18W for branch II, 8W for branch III E. 72W for branch I, 8W for branch II, 18W for branch III Solution. See derivation to the problem 15. The power released in the bulb of the branch I: ( i 1 ) 2 *R=E 2 /R=12*12/2=72 W The power released in each bulb (total two bulbs) of branch II: ( i 2 ) 2 *R= E 2 /(4R)= 12*12/4/2=18 W The power released in each bulb (total three bulbs) of branch III: ( i 3 ) 2 *R= E 2 /(9R)= 12*12/9/2=8 W Answer D. Problem 8. In the circuit shown in the figure find the currents through each resistor if E 1 =10 V, E 2 =5V and all resistors are equivalent with R 1 =R 2 =R 3 =R= 5Ω. A. i 1 = 1A down, i 2 =3A up, i 3 = 2A down

B. i 1 = 1A down, i 2 =1A up, i 3 = 0 C. i 1 = 2A down, i 2 =0, i 3 = 3A down D. i 1 = 3A up, i 2 =2A up, i 3 = 1A down E. i 1 = 1A up, i 2 =2A down, i 3 = 0 Solution. We have a real multi loop circuit problem now. We have three branches corresponding to three resistors. Inside each branch there is its own current. Lets enumerate currents in each resistors to be i 1, i 2, i 3 Assume that the direction of the current i1 in the resistor R1 is down, the direction of the current i2 in the resistor R2 is up and the direction of the current in the resistor i3 is down again. From the junction rule we obtain (i) i2=i1+i3 From two inner loops we obtain (ii) E1-i1*R-i2*R=0 (iii) E2-i3*R-i2*R=0 We have total three equations. Summing equations (ii) and (iii) and using Eq. (i): E1+E2-(i1+i3)*R-2*i2*R= E1+E2-3*R*i2= 0 Therefore i2=(e1+e2)/(3r)=15/15=1a (sign is +, therefore direction chosen is OK, i.e up) From equation (ii): i1=(e1-i2*r)/r=e1/r-i2=10/5-1=1a (sign is +, therefore direction of the current i2 is OK, i.e. down.) From equation (i): i3=i2-i1=1-1=0a (no current!) We finally have i1=1a down, i2=1a up, i3=0 Answer B.

9. RC CIRCUITS The RC circuit consists of the capacitor C, the resistor R, the battery E and the switch S. When the switch S is closed in RC circuit, capacitor is charged. Application of the Kirchoff loop equation for time moment t to RC circuit leads to differential equation. The solution of the equation reads as follows: (1.1) q t EC e ( ) = (1 ) is the charge on the capacitor as a function of time. Current in the RC circuit: (1.2) dq E E i t e e dt R R t / τc ( ) = = = where time constant τ C = RC shows the characteristic time for the charging process. The voltage across the capacitor can be found q( t) (1.3) V ( t) = = E(1 e ) C In the reverse situation, when the switch is opened, the capacitor with charge q0 gets discharged. The functions for the charge, current, and voltage are the following: (1.4) (1.5) q( t) = q e 0 dq q0 i( t) = = e dt RC q( t) q0 (1.6) V ( t) = = e C C Problem 9. What is the voltage across the capacitor in the RC circuit with R= 3 Ohm, C= 5 pf and E=9V just after the switch is closed. A. 0 V B. 1 V C. 3 V D. 5 V E. 9 V

Solution. After the switch is closed the capacitor begins to charge. The charge on the plates as a function of time is given by q t EC e ( ) = (1 ) The voltage across the capacitor is thus q( t) V t E e C ( ) = = (1 ) The time constant for the circuit RC is 5pF*3 Ohms = 15 picoseconds. Even though this is a very short time, it is not zero. Therefore, just after the switch is closed, t = 0 in the expression above makes the quantity in parentheses go to zero). Thus, the voltage V(t=0) = 0. Answer A. Problem 10. How long does it take for an RC circuit with R= 3 Ohm, C= 5 pf and E=9V to charge the capacitor to 45 pc. A. 1 second B. 1 hour C. 1 day D. 1 year E. infinity Solution. After the switch is closed the capacitor is beginning to charge. The charge on the plates as a function of time is: q t EC e ( ) = (1 ) We solve this for t with q known. Note that q is the limiting value = EC. t = RC q EC == = 12 ln 1 / 3*5*10 *ln 1 45/(9*5) inf Answer E

10. MAGNETIC FIELDS PLEASE NOTE THAT IN ALL FORMULAS THE VECTOR PRODUCT (CROSS PRODUCT) IS DENOTED WITH THE SQUARE BRACKETS: c=[axb] Key concepts: A moving charged particle creates a magnetic field around it. Do not confuse magnetic fields with electric fields: the electric fields are associated with the charges; the magnetic fields are associated with the moving charges. Therefore, a moving charged particle creates both an electric field and a magnetic field. If a moving charged particle is placed inside a magnetic field it experiences magnetic force. This force is proportional to the charge of the particle, its velocity and the strength of the magnetic field. Again, it is essential that the particle is moving: the faster the particle moves, the greater the force it feels from the external magnetic field. When a particle of charge q moves with velocity v inside a magnetic field B it feels a magnetic force F given by the formula: F =q*[vxb] which was found as the result of experimental measurement. A cross product [vxb] is a vector which is perpendicular to the plane made by vectors v and B. The direction of the cross product [vxb] is fixed by the right hand rule: rotate vector v towards vector B using your right hand (use the smallest angle between them). The orientation of your right hand shows the direction of the cross product. The direction of the magnetic force F can be different from the vector product [vxb] since the charge q may be negative. In that case the direction of the force F is opposite to the direction of the vector product since F =- q *[vxb]. Typical problems: Problem 11. Find the value and the direction of the force acting on a positively charged particle with q=2 µc moving with velocity v=5 km/s along x direction which is placed inside constant magnetic field B=1 kt oriented along y direction: A. 10 mn, -z direction B. 10 mn, +z direction C. 10 N, -z direction D. 10 N, +z direction E. 1 kn, +x direction

Solution. The vector product [vxb] is oriented along +z direction according to the right hand rule. Since the charge is positive the direction of the force is also lying along +z direction. The value for the force is (angle φ between v and B is 90 degrees) F=q*v*B* sinφ =q*v*b=2*10-6 *5*10 3 *1*10 3 =10 N Answer D Problem 12. Find the value and the direction of the force acting on a positively charged particle with q=2 mc moving with v=5 mm/s along x direction which is placed inside constant magnetic field B=1.41 kt oriented 45 degrees in xy plane A. 10 mn, -z direction B. 10 mn, + z direction C. 10 N, - z direction D. 10 N, +z direction E. 1 kn, +x direction Solution. The vector product [vxb] is oriented along +z direction according to the right hand rule. Since the charge is positive the direction of the force is also lying along +z direction. The value for the force is (angle φ between v and B is 45 degrees) F=q*v*B*sin φ =q*v*b sin45=2*10-3 *5*10-3 *1.41*10 3 /1.41=10 mn Answer B. Problem 13. Find the value and the direction of the force acting on a negatively charged article with q=-0.5 C moving with v=10 m/s along x direction which is placed inside constant magnetic field B=(-10,20,0) T Y y A. 10 N, -z axis B. 10 N, +z axis C. 100 N, -z axis D. 100 N, +z axis E. 1 N, x-axis Z z

Solution. Since vector B has z-component equal zero, it is completely oriented within xy plane. Therefore, the vector product [vxb] is oriented along +z direction according to the right-hand rule. Since the charge is negative the direction of the force is lying along -z direction. Let us find the direction and the value for the force using vector algebra. First, let us represent the vector B in the form: B=-10* i +20*j, where i is a unit vector along x axis and j is a unit vector along y-axis. Vector v=10*i is lying completely along x axis. The vector product is: [vxb]= [ 10*i*(-10* i +20*j)]=-10*10*[i*i]+10*20*[i*j] Since vector product [ixi]=0 and [ixj]=k, the unit vector along z-axis, the vector product becomes: [vxb]=200* k. Now the force is the vector product time the charge. Since the charge is negative: F=q*[vxB]=-0.5*200* k=-100*k, N, i.e 100 N oriented along vector k which is z axis. Answer C. Problem 14. An electron moving with the constant speed in the direction from top to bottom enters the region of a constant magnetic field. The electron trajectory deviates to the left. What is the orientation of the magnetic field? A. top B. bottom C. left D. right E. inside the page Solution. If electron deviates to the left, it experiences the force directed to the left. This is the magnetic force described as F =q*[vxb]. We need to figure out such direction of B so that to be multiplied by vector v and taking account the sign of charge for the electron, we obtain the direction of the force to the left. To do this, we first note that if the force is directed to the left, the vector product [vxb] should be directed to the right since electron is negatively charged particle (q<0). Second, if v is directed to the bottom, vector B should be directed inside the page, in this way vector product [vxb] is directed to the right, and the force will be directed to the left. Answer E. Problem 15. An electron is circulating within the plane of the page due to magnetic field directed inside the page. The radius of its circulating motion is 9 cm and its speed is 1.6 km/s. Find the value of magnetic field and the direction of the circulating motion,

A. 0.1 µt, clockwise B. 0.1 µt, counterclockwise C. 0.1 mt, clockwise D. 0.1 mt, counterclockwise E. 0.1 T, clockwise Solution. Circular motion assumes that there is force acting on the electron, which is directed towards the center of the orbit. This force is of magnetic origin, i.e. F =e*[vxb]. At the same time, the force associated with the circular motion is equal to m*w where centrifugal acceleration w= v 2 /R. Since velocity vector v is always perpendicular to magnetic field vector B, we can write F=e*v*B=m*v 2 /R from which B=m*v/(R*e)=9*10-31 *1.6*10 3 /(9*10-2 *1.6*10-19 )=10-7 T. To understand whether the electron circulates clockwise or counterclockwise, we need to look at the electron at some of its position. Assume that at some moment t it is positioned at 12pm (the topmost part of its trajectory). If its velocity is directed to the right, it is going clockwise, if its velocity is directed to the left, it is going counterclockwise, At the topmost point of its trajectory the electron should experience the force directed to the bottom since it is going around the circle, and the force of the circular motion is always directed to the center. Therefore F =e*[vxb] is directed to the bottom, and we know that e<0 and B is directed inside the page. The vector product [vxb] should be directed to the top since F is directed to the bottom, and e is less than zero. If vector product is directed to the top and B is directed inside the page, the velocity vector v should be directed to the right. In this way, vector product [vxb] will be directed to the top, and multiplied by negative charge, the force will be directed to the bottom. We therefore found that the velocity in topmost point is directed to the right. Therefore, the electron is going around the circle clockwise. We can fix any other of its position (say leftmost) and come to the same conclusion. Answer A.

11. MAGNETIC FIELDS FROM CURRENTS Key concepts: In general, the magnetic field due to a moving charged particle is given by: B(r)= µ 0 /4π *q*[v*r]/r 3 where constant µ 0 /4π=10-7. Note that the direction of the magnetic field is tricky: if the particle is moving from bottom to top, and a point where we look at the field is on the right, the vector product [vxr] is directed inside the page. If the charge of the particle >0, the magnetic field will be pointed inside the page at this point r. If the charge < 0 the magnetic field is pointed outside the page at this point r. All this information is encoded into the above formula. Since current inside the wire is just an array of moving charged particles, every wire carrying current creates magnetic field around it. To find the total field at some point r we need to sum up all the contributions to this field from all moving charges inside the wire. In general, there is an integral, which gives the answer. For simple configurations like straight wire or circular wire the answers are: 1. Infinite straight wire carrying current creates magnetic field which varies with the distance d from the wire as follows: B(d)=µ 0 *i/(2π d), where i is the value of the current. The direction of the field can be found using the right hand rule: point your large finger along with the current, the four other fingers show the direction of the magnetic field (See figure in the textbook). 2. A coil or a circular wire creates magnetic field at the center of the circle: B= µ 0 *i /(2R). The direction of the field can be found from right hand rule: rotate by your right hand in the direction of the current simulating circular motion, the orientation of the right hand will show the direction of the magnetic field within the plane inside the circle. A general arc defined by angle ϕ creates magnetic field B=µ 0 *I*ϕ /(4πR) which for ϕ=2π transforms to the expression above. 3. A solenoid, which is a cylindrically turned wire with some (usually large) number of turns per unit length, creates non-zero constant magnetic field inside itself and zero field outside itself. The magnetic field inside the solenoid is given by: B=µ 0 *n*i, where n is the number of turns per unit length. The direction of the field inside the solenoid can be found by applying the right hand rule: rotate by your right hand in the direction of the current simulating circular motion, the orientation of the right hand shows the direction of the field inside the solenoid. A wire with the current placed into magnetic field experiences a force since every moving charge inside the wire experiences the magnetic force. The total force on the wire with current i depends on the orientation of the wire and on the orientation of the magnetic field. If the wire is a straight piece of some length L, the force is given by F= i* [ L xb], where vector L has length L an is directed towards the direction of the current.

Two pieces of wire, which brought together will repel from or attract to each other if the currents exist inside these wires. This can be simply understood since one wire will create magnetic field around it, and the moving particles inside another wire will experience the magnetic force. For example, if two parallel wires carry currents in the same direction, they will attract to each other. If two parallel wires carry currents in different direction, they will repel from each other. In case of two parallel pieces of wire of length L separated by distance d which carry out the currents i1 and i2, the force between them is given by F=µ 0 i1*i2*l/(2πd). Typical problems: Problem 16. Find the direction and the value for the force on the wire, which carries the current of 2 ma, if it is placed inside the magnetic field of 4 kt. The length of the wire is 3 mm, it is vertically oriented with the current flowing to the bottom. The magnetic field is horizontally oriented and is pointed to the left. A. 24 Ν, left B. 24 mν, right C. 24 kn, top D. 24 N, bottom E. 24 mn, inside the page Solution. The formula to use is: F= i*[ L xb], where L is a vector which points from top to bottom (vertical orientation with the current flowing to the bottom) and B points to the left. Both vectors are perpendicular to each other. The vector product [LxB] points inside the page. We should not care about sign of i in this case, it is always positive, and direction of vector L takes care about direction of the current. The value of the force is F=i*L*B=2*10-3 *3*10-3 *4*10 3 =24 mn. Answer E. Problem 17. A wire carries a current from the right to the left. What is the orientation of the magnetic field above and below the wire? A. Above inside the page, Below outside the page B. Above outside the page, Below inside the page C. Above left, Below right D. Above right, Below left E. Above bottom, Below top

Solution. Applying the right hand rule, orient your large finger to the left, the four other fingers point to the direction of the field above the wire, they point to inside the page. If you rotate by the right hand around the wire so that your large fingers are located below the wire, the point outside the page. This gives the answer to the problem. Answer A. Problem 18. Find the value of magnetic field of the long straight wire at distance d=3 cm if it carries current of 3 ma. A. 20 pt B. 20 nt C. 20 µt D. 20 mt E. 20 T Solution. The formula is B(d)= µ 0 *i/(2π d), where all numbers are given by the problem. B(d)= µ 0 *i/(2π d)= 2 µ 0 *i/(4π d)=2*10-7 *3*10-3 /(3*10-2 )=2*10-8 =20 * 10-9 T=20 nt Answer B. Problem 19. Two straight wires of the length 5 m each and parallel to each other carry currents of 3 A and 5 A in opposite directions. If the distance between the wires is 1 cm what is the force between them? A. 3 mn, repulsion B. 3 mn, attraction C. 1.5 mn, repulsion D. 1.5 mn, attraction E. 3 N, repulsion Solution. The formula to use is: F=µ 0 i1*i2*l/(2πd)= 2* 10-7 *3*5*5/0.01=150*10-5 N. Since the directions of the currents are opposite, it is repulsion. Answer C.

Problem 20. A circular wire of radius 6.28 cm is oriented within the page and carries a current of 2 A. The current flows clockwise. What are the value and the direction for the magnetic field created at its center? A. 20 µt, outside the page B. 20 µt, inside the page C. 20 mt, to the left D. 20 mt to the right E. 20 T, to the top Solution. The formula to use is: B=µ 0 *i /(2R)=4*3.14 * 10-7 * 2 / (2 * 6.28 * 10-2 )= 2*10-5 T. Let us apply the right hand rule: to simulate the current flow by rotating the right hand, the orientation of the right hand should be inside the page. Answer B. Problem 21. A solenoid is oriented perpendicular to the page, and carries a current in the counterclockwise direction. What are the value and the direction of the magnetic field inside it? Assume that the current is 1 A, and the number of turns per centimeter is 25. A. 3.14 mt, outside the page B. 3.14 mt, inside the page C. 3.14 T, to the left D. 3.14 T to the right E. 3.14 kt, to the top Solution. The formula to use is: B=µ 0 *i*n=4*3.14*10-7 *1*25*10 2 =3.14 mt. The direction is outside the page, since simulation of the current flow by rotating your right hand gives us the answer. Answer A.

Problem 22. Find the direction of the force on each side of rectangular frame. A. Top side - top, bottom side - bottom, left side - left, right side right. B. Top side - bottom, bottom side - top, left side - right, right side left. C. Top side - left, bottom side - right, left side - bottom, right side top. D. Top side - right, bottom side - left, left side - bottom, right side top. E. Top side - left, bottom side - right, left side - top, right side bottom. Solution. Parallel wires attract to each other if the currents flow in the same direction. They repel from each other if the currents flow in opposite direction. For perpendicular wires the situation is more complicated. It can be figured out by noting that the wire below itself creates a magnetic field directed inside the page. Since the current flow in the left side of the frame is on top, this is the flow of positively charged particles. They will experience the force F =q*[vxb], pointed to the left (q>0, v is on top, B is inside the page). Therefore the entire left part of the frame will experience the force pointed to the left. Using similar argument, the entire right part experiences the force pointed to the right. Answer A. Problem 23. Find the total force on the frame in the above problem if the current inside the wire is 30 A, and inside the frame is 20 A. The horizontal dimension of the frame is 30 cm and the vertical dimension is 7 cm. The top side of the frame is located at the distance 1 cm from the wire. A. 3.15 N, top B. 3.15 N, bottom C. 3.15 mn, top D. 3.15 mn, bottom E. 3.15 kn, left Solution. First note that the forces on left side and right side are the same and directed oppositely. Therefore they do not contribute to the total force. The force on the top side is directed to the top and it is given by: F top =µ 0 *i1*i2*l/(2πd) with d=1cm. Evaluating it: 2*10-7 *30*20*30*10-2 /10-2 =36*10-4 N. The force on the bottom part is directed to the bottom and it is given by the same formula with d = 8 cm. Therefore, the force F bottom =µ 0 *i1*i2*l/(2πd)= 2*10-7 *30*20*30*10-2 /(8*10-2 ) = 4.5*10-4 N. The total force is directed on top and it is 36*10-4 -4.5*10-4 =31.5*10-4 N.= 3.15 mn.