MAGNETIC FIELDS AND FORCES

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Juniper Carpenter
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1 PHYSICS 120 : ELECTRICITY AND MAGNETISM TUTORIAL QUESTIONS MAGNETIC IELDS AND ORCES Question 96 A charge of 12.0 ñc, travelling with a speed of ms 1 in a direction perpendicular to a magnetic field, experiences a magnetic force of N. Calculate the magnitude of the magnetic field. The magnitudes of the magnetic field B and the magnetic force are related by = q vbsinθ B = q vsinθ = ( ) ( ) sin90 = T Question 97 A particle, moving with a velocity of ms 1 at an angle of 30 with respect to a magnetic field of T, experiences a force of N. Calculate the magnitude of the particle s charge. = q vbsinθ q = vbsinθ = ( ) ( ) sin30 = C page 1 of 7

2 Question 98 Particles 1, 2 and 3 follow the paths shown below as they pass through a magnetic field. What can one conclude about each particle? Since particle 2 s trajectory is unchanged, it must be a neutral particle. The paths of particles 1 and 3 is affected by the magnetic field, and by the right-hand rule it can be seen that 1 must be a negatively charged particle, while 3 is positively charged. Question 99 Three particles have identical charges and masses. They enter a uniform magnetic field and follow the paths shown below. Which particle is moving the fastest and which is moving the slowest? Justify your answers. The force is given by = q vbsinθ and since q, B and sinθ are the same for each particle, then v. Also, the resultant force in circular motion is = v2 r so that v r, since the particles have the same mass. Particle 2 s trajectory has the smallest r and so it has the smallest v. Particle 1 has the largest v. Question 100 A wire 1.0m long carries a current of 10A and makes an angle of 30 with a uniform magnetic field with B = 1.5T. Calculate the magnitude and direction of the force on the wire. is perpendicular to B and the wire. = BIlsinθ = (1.5) (10) (1) sin30 = 7.5N page 2 of 7

3 Question 101 (a) Calculate the force per unit length on a wire carrying a current of 0.50A when perpendicular to a 4.0T magnetic field. (b) What if the angle between the wire and the field is 45? (a) l (b) When the angle is 45 l = BIsinθ = (4.0) (0.50) sin90 = 2.0Nm 1 = BIsinθ = (4.0) (0.50) sin45 = = 2Nm 1 Question 102 The force on a wire carrying 20A is a maximum of 3.6N when placed between the pole faces of a magnet. If the wire is 15cm long, what is the approximate magnitude of B? The maximum force implies that the angle is 90. The magnitude of B is B = Il = = 1.2T Question 103 How far from a long, straight wire carrying 10A will the magnetic field be T? B = µ 0 I 2π r r = µ 0 I 2πB = = m Question 104 Determine the magnetic field midway between two long straight wires 10 cm apart if one carries 10A and the other 8.0A and these currents are (a) in the same direction, and (b) in opposite directions. (a) The magnetic field is additive. Midway between the wires, the fields are either in the same direction or in the opposite direction. Let both wires carry current out of the plane of the paper. page 3 of 7

4 The fields are in opposite directions. Hence B = B A B B = µ 0 2π (I A I B ) = (10 8.0) = T SinceB A wastakenaspositive, theresultantmagneticfieldpointsinthesamedirection B A. (b) or currents in opposite directions, the field midway points in the same direction for boh wires. B = B A +B B = µ 0 2π (I A +I B ) = T Question 106 Calculate the magnitude and direction of the force between two parallel wires 85.0 m long and 30.0cm apart, each carrying 60.0A in the same direction. The force on a current carrying wire in a magnetic field is = BIlsinθ. Here, magnetic field is due to the other wire, and the angle is 90. Hence ( ) µ0 I 1 = I 2 l = ( ) = 0.204N 2π r The wires carry current in the same direction so the force is attractive. Question 107 A vertical, straight wire carrying 5.0A exerts an attractive force per unit length Nm 1 on a second, parallel wire 8.0cm away. What current (magnitude and direction) flows in the second wire? The force is attractive therefore the second wire carries current in the same direction as the first. = µ 0 I 1 I 2 l 2π r I 2 = r 2π l I 1 µ 0 = ( ) = 48A ( ) 5.0 page 4 of 7

5 Question 109 A solenoid 120cm long and 3.0cm in mean diameter has five layers of windings with 840 turns on each. The current in the solenoid is 5.0A. Calculate (i) the magnetic field at the centre of the solenoid and (ii) the magnetic flux per turn for a cross-section of the solenoid at its centre. (i) The total number of turns is The magnetic field is B = µ 0 ni = 4π 10 7 (ii) The total flux inside the solenoid is N = = = T φ = BAN cosθ so the flux per turn is ( φ N = BAcosθ = B(πr2 )cosθ = ( ) π ( ) ) cos0 = Wb/turn 2 Question 112 A closely-wound rectangular coil of 50 turns has dimensions of 12.0cm 25.0cm. The plane of the coil is rotated in time t = 0.100s from a position where it makes an angle of 60.0 with a uniform magnetic field of strength 2.00T to a position perpendicular to the field. What is the average emf induced in the coil? Let the average emf E be dφ E = = φ The component of the field perpendicular to the area A goes from Acos30 to Acos0 in 0.10s. Thus the average emf is given by E = φ = BAN(cosφ cos I ) = 2.00 ( ) ( ) (50.0) (cos0 cos30 ) = 4.02V page 5 of 7

6 Question 113 A rectangular coil having 10 turns with dimensions of 20.0cm 30.0cm rotates with an angular velocity of 600rpm in a uniform magnetic field of strength 0.100T. The axis of rotation is perpendicular to the field. ind the maximum emf produced. With a diagram similar to Question 112, here B, A and N are constants. The magnitude of the induced emf is E = dφ = d (BAN cosθ) = BAN d (cosθ) = BAN dcosθ Now dcosθ = sinθ and can have a maximum value of 1. Also is the angular frequency ω, where The maximum emf is then ω = 2πf = 2π(600)rad/min = 20πrad/s E MAX = BANω = (0.100) ( ) ( ) (10) (20π) = 3.77V Question 115 A closely-wound coil has an area of 4.0cm 2, 160 turns and a resisntance of 50Ω. It is connected to a charge-measuring instrument whose resistance is 30 Ω. When the coil is rotated quickly from a position parallel to a uniform magnetic field to one perpendicular to the field, the instrument indicates a charge of C/ What is the magnitude of the field? Ignoring the reactance of the coil, the average current I is E = q = A The average potential difference for the coil and charge-measuring instrument is V = I R = (50+30) = page 6 of 7

7 The magnitude of the average emf is PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS E = B = φ = BAN (cos0 cos90 ) = BAN (cos0 cos90 ) = T page 7 of 7

Phys222 Winter 2012 Quiz 4 Chapters 29-31. Name

Phys222 Winter 2012 Quiz 4 Chapters 29-31. Name Name If you think that no correct answer is provided, give your answer, state your reasoning briefly; append additional sheet of paper if necessary. 1. A particle (q = 5.0 nc, m = 3.0 µg) moves in a region