# PHYS 155: Final Tutorial

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1 Final Tutorial Saskatoon Engineering Students Society April 13, 2015

2 Overview

3 Tutorial Slides These slides have been posted: sess.usask.ca homepage.usask.ca/esp991/

4 Section 1

5 have no start, no finish. There is NO point of origin. They are created by movement of charge. B = 0 B = µ 0 J + µ 0 E 0 de dt Changing fields can incite EMF in other objects. E = db dt

6 Field Lines Field lines have no start, no finish. field strength B related to density of field lines.

7 Three Very Spatial Assume Positive Charge and Current Find magnetic field direction in various cases Find force on a moving charge or current carrying wire.

8 RHR 1: Field Around a field circles around wires, in a direction CCW when wire is coming out of page. Put your thumb along wire. Your fingers curl around wire along magnetic field lines.

9 RHR 2: Field Inside a loop or Coil of wound in a coil generates a magnetic field directed along the axis of the coil. Fingers wrap around the loop or coil, in the direction of the current. Thumb points in the direction of magnetic field.

10 RHR 2: Additional Consideration We can find RHR No. 2 by applying Rule No. 1 to a coil instead of a straight wire!

11 RHR 3: Force on a Moving Charge Charge or current moving in a magnetic field experiences a magnetic force. Perpendicular to direction of movement Perpendicular to magnetic field.

12 RHR 3: Force on a Moving Charge Thumb Along Direction of Movement Index Finger along Field Palm / middle finger points to the force! Remember, this is for POSITIVE Charge. For negative charge like electrons, the force is opposite!

13 RHR 3: Force on a Moving Charge

14 : RHR #1: Straight Thumb Along Direction of Current Fingers curl the way the magnetic field does. RHR #2: Coil / Loop of Curl fingers around coil of wire in direction of current Thumb tells you direction of magnetic field RHR #3: Charge Moving in Field Thumb Along Movement of Positive Charge Index finger along magnetic field direction Palm or middle finger tells you direction of force on positive charge.

15 Some Helpful Formulas Field for a long, straight wire Field at Centre of Loop Field Inside Solenoid B = µ 0I 2πr B = N µ 0I 2R B = µ o ni = µ 0 N l I Force on Moving Charge / in Field F = q(v B) F = qvb sin θ

16 Example 1 Lorentz Force A proton moving at m/s through a magnetic field of magnitude 1.70 T experiences a magnetic force of magnitude N. What is the angle between the proton s velocity and the field?

17 Example 1 Solution Draw a picture showing this effect. Lorentz force is so F = q v B F = qvb sin θ Using magnitude of force, velocity and magnetic field, solve for θ. Answer: θ = or

18 Example 2 Cyclotron Motion A singly charged ion of mass m is accelerated from rest by a potential difference V. It is then deflected by a uniform magnetic field (perpendicular to the ion s velocity) into a semicircle of radius R. Now a doubly charged ion of mass m is accelerated through the same potential difference and deflected by the same magnetic field into a semicircle of radius R = 2R. What is the ratio of the masses of the ions?

19 Example 2 Solution Write the kinetic energy of the particle. KE = q V = 1 2 mv 2 (1) Write the conditition for centripetal acceleration F m = F c, qvb = mv 2 R (2)

20 Example 2 Solution Sub (1) into (2) and solve for m in terms of q, B, V and R. m = B2 2 V qr2 Solve for m in terms of q, B, V and R. Note that all you have to do is tweak your q, B & R values. Answer: m = 8.00 m. m = B2 2 V (2q)(2R)2

21 Example 3 Hold Straight Condition A velocity selector consists of electric and magnetic fields described by the expressions E = E ˆk and B = Bĵ, with B = 15.0mT. Find the value of E such that a 750-eV electron moving in the negative x direction is undeflected.

22 Example 3 Solution Draw your picture. Which way will the magnetic field try to direct the electron? (Up) Which way does the electric force need to apply? (Down). Which way should the field be pointed? (Up) Solve for electron velocity using K E = 1 2 mv 2 Draw FBD and write the condition for no deflection: F M + F E = 0

23 Example 3 Solution Figure out the magnitude of the magnetic force on the electron. Note that Solve for E. Answer: 244 kv/m F M = q v B = ˆk N F E = E ˆk = F M

24 Break See you in 10 Minutes

25 Section 4 Torque on due to Field

26 Force on a Extension of RHR #3. F = I ( l B) F = I lb sin θ l is length of wire, whose direction is the direction of the current. For 2 parallel wires, the force of attraction/repulsion is F = µ 0 I 1 I 2 2π d L Attraction if current in same direction Repulsion if current in opposite direction

27 Torque on a Loop A loop of wire in a magnetic field expriences a torque, according to τ = NI [ A B] τ = NIAB sin θ the A vector has magnitude equal to area of loop, direction points in direction of magnetic field generated by loop (use RHR 2).

28 Example 4 Torque on A current of 17.0mA is maintained in a single circular loop of 2.00 m circumference. A magnetic field of T is directed parallel to the plane of the loop. What is the torque exerted on the loop by the magnetic field?

29 Example 4 Solution 1 Draw a picture of the loop of wire and the field. Conceptualize: if current flows in the loop, and the magnetic field is parallel to the plane of the loop, which side will try to lift up? Which side will be pushed down? 2 Solve for area of the loop using circumference. 3 Use Torque Formula to solve for torque. Answer: N m. τ = NI [ A B]

30 Example 5 Attraction Between s The current in the long, straight wire is I 1 = 5.00 A and the wire lies in the plane of the rectangular loop, which carries a current I 2 = 10.0 A. The dimensions in the figure are c =0.100 m, a = m, and l = m. Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire.

31 Example 5 Solution 1 Ask yourself: Which way will each part of the loop try to move? 2 Find the strength of the magnetic field due to the long wire at the left side of the loop. B = µ 0I 1 2πc 3 Find the Lorentz force on the left side of the loop F = I 2 l B 4 Find the magnetic field strength and the force on the right side of the loop. B = µ 0 I 1 2π(c + a) F = I 2 l B

32 Example 5 Solution Note that the top and bottom parts of the loop experience forces too, but they cancel out. Subtract the force of the right side of the loop from the force of the left side of the loop to find net force. Answer: î N

33 Section 5

34 Let s us find the magnetic field around an enclosed loop of wire. Comes from B = µ 0 J + µ 0 E 0 de dt This simplifies to B dl = µ0 I enc. Pick some sort of symmetry, usually circular, around some source of current. Then you can find the magnetic field strength.

35 Example 6 A packed bundle of 100 long, straight, insulated wires forms a cylinder of radius R = cm. If each wire carries 2.00 A, what are the (a) magnitude and (b) direction of the magnetic force per unit length acting on a wire located cm from the centre of the bundle?

36 Example 6 Solution 1 Recall : B dl = µ 0 I enc 2 Figure out the current enclosed by a radius of 0.2cm. (It s 32 amps). 3 If we integrate around a circle, centred at the centre of the wire, B will be constant because of radial symmetry. So you get 4 Solve for B B (2πr) = µ 0 I enc 5 Multiply by 2A, the current of 1 wire at that radius. (F/l=IB) Answer: F/l = 6.4 mn.

37 Break See you in 10 Minutes

38 Section 7 and Lenz s Law

39 Lenz s Law A loop or coil of wire that is exposed to a change in magnetic flux through the loop will induce an EMF whose corresponding magnetic field will oppose the original change in flux.

40 Mathematical Representation of Lenz s Law E = N dφ dt = N d (AB cos θ) dt So we get Back EMF for one of three reasons: Change in Area, Field Strength or Angle. Usually only one will change at a time in any given question.

41 Example 7 Faraday Law A coil formed by wrapping 50 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of 30.0 with the direction of the field. When the magnetic field is increased uniformly from 200 µt to 600 µt in s, an emf of magnitude 80.0 mv is induced in the coil. What is the total length of the wire in the coil?

42 Example 7 Solution 1 Recall Faraday Law E = N dφ dt = N d (AB cos θ) dt 2 Here, only the magnetic field is changing, so this simplifies to E = NA db dt cos θ 3 Solve for A, then find side length, then multiply by 4N to get length of wire. Answer: 272m.

43 Bar and Rail Problems 1 Figure out the induced EMF Lenz s Law to find direction. for magnitude. 2 Using EMF, find current in loop (Ohm s Law, KVL) 3 Using I and Lorentz force law, find magnetic force on bar. 4 Draw FBD of Bar, solve unknowns. Find any input force required. 5 Watch out for directions of B, v and F, especially on angled rails!

44 Example 8 Bar and Rail The picture shows a bar of mass m = kg that can slide without friction on a pair of rails separated by a distance ` = 1.20 m and located on an inclined plane that makes an angle θ = 25.0 with respect to the horizontal. The resistance of the resistor is R = 1.00 Ω and a uniform magnetic field of magnitude B = T is directed downward, perpendicular to the ground, over the entire region through which the bar moves. With what constant speed v does the bar slide along the rails?

45 Example 8 Solution 1 Solve for EMF using E = lvb cos θ 2 Use Ohm s Law to find current through loop. I = E/R 3 Lorenz Force Law (F = qvb) to find the back EMF force on the bar (* Watch out for direction here *). 4 Draw FBD to find relation between F M, N and F g. 5 For constant velocity, you can relate F M and F g. Solve for v Answer: 2.80 m/s v = mg tan θr l 2 B 2 cos θ

46 Section 8

47 Self E = N dφ dt Mutual E L = L di dt L = µ 0 N 2 l A = µ 0µ r N 2 A l Energy In an Inductor N s Φ s = MI P U = 1 2 LI 2 E s = M di P dt

48 Example 9 Mutual Two solenoids A and B, spaced close to each other and sharing the same cylindrical axis, have 400 and 700 turns respectively. A current of 3.50 A in solenoid A produces an average flux of 300 µwb through each turn of A and a flux of 90.0 µwb through each turn of B. (a) Calculate the mutual inductance of the two solenoids. (b) What is the inductance of A? (c) What emf is induced in B when the current in A changes at a rate of 0.5 A/s?

49 Example 9 Solution 1 Use the mutual inductance formula that has all the variables you know. N S Φ S = MI P 2 To find self inductance, use self inductance formula relating flux and current. NΦ = LI 3 To find EMF, use EMF formula with mutual inductance. E S = M di P dt

50 Example 9 Solution It might be helpful to pay attention to the directions of the coils here. Remember that the - serves to remind you that it is back EMF, but the winding of the coil actually decides the direction of current. Answer: H, H, -9mV.

51 Transformers Just remeber the transformer turn ratio formula, relating current, voltage and number of turns in the Primary and Secondary coils. N P N S = V P V S = I S I P Use Ohm s Law and P = VI to solve for everything else.

52 Example 10 Transformers In the transformer shown, the load resistance R L is 50.0 Ω. The turns ratio N 1 /N 2 is 2.50, and the rms source voltage is V s = 80.0 V. If a voltmeter across the load resistance measures an rms voltage of 25.0 V, what is the source resistance R s?

53 Example 10 Solution Remember the big important transformer equation. N P N S = V P V S = I S I P Using the secondary coil voltage and Ohm s law, find current in secondary coil. Using turns ratio, find the current in primary coil and voltage in primary across the transformer. KVL to find voltage across resistor, and Ohm s Law to find resistance. Answer: 87.5 Ω

54 1 Go through the slides (both mine and the ones on blackboard) and make up a solid formula sheet. 2 Have a good section on unit conversions, and what units you can expect from certain formulas. 3 Do the assignments. Each question has a recipe to solve it. Practise your Lenz Law to anticipate direction of magnetic fields and back EMF. 4 Watch out for weird units on the exam.

55 Good Luck! These slides have been posted: sess.usask.ca homepage.usask.ca/esp991/

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