mv = ev ebr Application: circular motion of moving ions In a uniform magnetic field: The mass spectrometer KE=PE magnitude of electron charge


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1 1.4 The Mass Spectrometer Application: circular motion of moving ions In a uniform magnetic field: The mass spectrometer mv r qb mv eb magnitude of electron charge 1 mv ev KEPE v 1 mv ebr m v e r m B m e r m ev er V B B The mass spectrum of naturally occurring neon, showing three isotopes. 1
2 Example: A singly charged positive ion has a mass of.5 x 106 kg. After being accelerated through a potential difference of 50 V, the ion enters a magnetic field of 0.5 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field.
3 Example: A singly charged positive ion has a mass of.5 x 106 kg. After being accelerated through a potential difference of 50 V, the ion enters a magnetic field of 0.5 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field. q 1.6x10 19 m.5x10 6 V 50 V B 0.5 T r? C kg mv FB Fc qvb r r V v W q K q 1 mv q mv qb We need to solve for the velocity! 19 Vq (50)(1.6x10 ) m.5x10 m/s r (.5x10 (1.6x10 6 )(56,568) 19 )(0.5) m 3
4 1.3 The Motion of a Charged Particle in a Magnetic Field Example: Velocity Selector A velocity selector is a device for measuring the velocity of a charged particle. The device operates by applying electric and magnetic forces to the particle in such a way that these forces balance. Given B and q, wow should an electric field be applied so that the force it applies to the particle can balance the magnetic force? Solution: by RHR1: the velocity is to the right (thumb) and the magnetic field (finger) is into the page ( x marks the tail of an arrow), so the magnetic force on a positive charge is upward (palm up). Here θ 90 F B qvb We therefore need the electric force F E that points down that has the same magnitude: for a positive charge the electric field then needs to point in the same direction as the desired force, and F E qe. We want F E F B qe qvb E vb (and pointing downward) 4
5 1.5 The Force on a Current in a Magnetic Field Magnetic force on a current The magnetic force on the moving charges pushes the wire to the right. Since a current consists of moving charges then a current is subject to the Lorentz force Magnetic force on a straight current segment of length L F qvbsinθ In Class Demo F q t L I ( v t ) Bsinθ F ILBsinθ Here θ is the angle between the direction of the current and the magnetic field 5
6 1.5 The Force on a Current in a Magnetic Field Example: A wire carries a current of.0 A from west to east. Assume that at this location the magnetic field of Earth is horizontal and directed from south to north and that it has a magnitude of B T. (a) Find the magnitude and direction of the magnetic force on a 36.0 m length of wire. (b) Calculate the gravitational force on the same length of wire if it s made of copper and has a crosssectional area of m. 6
7 1.5 The Force on a Current in a Magnetic Field Example: A wire carries a current of.0 A from west to east. Assume that at this location the magnetic field of Earth is horizontal and directed from south to north and that it has a magnitude of B T. (a) Find the magnitude and direction of the magnetic force on a 36.0 m length of wire. (b) Calculate the gravitational force on the same length of wire if it s made of copper and has a crosssectional area of m. Solution : (a) By RHR 1 the force is pointing upward F ILB sinθ B whereθ is the angle between the direction of the current segment and the magneticfield Here θ 90 F (b) Density of copper is ρ Mass of wire : M ρ( volume) ρal M F F G G ( Mg / F B 3 ( kg)(9.81m/s 199 B kg/m ILB 3 (.0 A)(36.0 m)( )( m kg/m ) 7.89 N 35 )(36.0 m) kg T) N (up) 7
8 1.7 Magnetic Fields Produced by Currents In Class Demo A LONG ( ), STRAIGHT WIRE carrying current produces a magnetic field µ B oi π r Result to remember: derivation requires integral calculus µ o 4π 10 7 T m A Magnitude permeability of free space Direction RightHand Rule No.. Curl the fingers of the right hand into the shape of a halfcircle. Point the thumb in the direction of the conventional current, and the tips of the fingers will point in the direction of the magnetic field. 8
9 1.7 Magnetic Fields Produced by Currents Current carrying wires can exert forces on each other. Example: What is the attraction/repulsion force per meter of wire between two infinite straight wires carrying 1.00A each, separated by 1.00 m? 9
10 1.7 Magnetic Fields Produced by Currents Current carrying wires can exert forces on each other. Example: What is the attraction/repulsion force per meter of wire between two infinite straight wires carrying 1.00A each, separated by 1.00 m? B 1 F F L µ 0I1 (4π 10 T m/a)(1.00 A) π r π (1.00 m) I LB1 sinθ, θ 90 µ 0I1I I B1 π r 7 (1.00 A)( T) 7 N s C C m s N m T 1 m This is in fact the original SI definition of an ampere (A). 10
11 1.7 Magnetic Fields Produced by Currents SOLENOID number of turns per unit length n N / L Magnetic field in the interior of a solenoid Is approximately uniform In the limit of infinite length, but finite n B µ ni o Magnitude µ oni L Result to remember: derivation requires integral calculus 11
12 1.7 Magnetic Fields Produced by Currents A CIRCULAR LOOP OF WIRE R B R.H. Two different applications of RHR for the same field configuration At the center of the circular loop only Magnitude µ B oi R Result to remember: derivation requires integral calculus 1
13 1.7 Magnetic Fields Produced by Currents Example: Net Magnetic Field A long straight wire carries a current of 8.0 A and a circular loop of wire carries a current of.0 A and has a radius of m. Find the magnitude and direction of the magnetic field at the center of the loop. 13
14 1.7 Magnetic Fields Produced by Currents Example: Net Magnetic Field A long straight wire carries a current of 8.0 A and a circular loop of wire carries a current of.0 A and has a radius of m. Find the magnitude and direction of the magnetic field at the center of the loop. B µ I 1 µ oi µ o I1 π r R π r I R o B ( 7 π 10 T m A) A.0 A T π ( m) m 14
15 1.7 Magnetic Fields Produced by Currents The field lines around a current loop resemble those around the bar magnet. Attraction Repulsion 15
16 1.6 The Torque on a CurrentCarrying Coil Put a rectangular loop of current I and length (height) L, and width w in a uniform magnetic field B. The loop is mounted such that it is free to rotate about a vertical axis through its center. We will consider the forces on each segment and the resulting torque from each. Using RHR1: The force on segment 3 points down, and that on segment 4 points up. F 3 and F 4 are also equal in magnitude and cancel one another. The magnitudes F 3 F 4 IwBsin(90 φ) IwBcosφ also change with the rotation angle φ But both F 3 and F 4 are directed parallel to the axis, and results in no torque. Top view φ is the angle between the normal to the loop and the magnetic field 16
17 1.6 The Torque on a CurrentCarrying Coil Looking at segments 1 and which have the current running vertically. By RHR1, force F 1 on segment 1 (current up) points into the page, for all values of φ. Also by RHR1, force F on segment 1 (current down) points out of the page. They cancel each other to yield no net force on the loop. However, F 1 and F both tend to turn the loop in the clockwise sense (as seen in the top view). The torques from the two forces are each w τ 1, ( F1, ) sinφ F ILB sinθ sinceθ 90 τ τ 1 + τ Fwsinφ ILB Top view 1 φ is the angle between the normal to the loop and the magnetic field 17
18 1.6 The Torque on a CurrentCarrying Coil τ τ 1 + τ Fwsinφ The torque τ is maximum when the normal of the loop is perpendicular to the magnetic field, and zero when the normal is parallel to the field. F ILB sinθ sinceθ 90 ILB The torque tends to cause the loop normal to become aligned to the field, just like on a bar magnet. Current loop magnetic dipole ( wsinφ) IAB sinφ Net torque τ ILB Top view A Lw area of loop τ magnetic moment m NIA B sinφ number of turns of wire 18
19 The Torque on a CurrentCarrying Coil Example The Torque Exerted on a CurrentCarrying Coil A coil of wire has an area of.0x104 m, consists of 100 loops or turns, and contains a current of A. The coil is placed in a uniform magnetic field of magnitude 0.15 T. (a) Determine the magnetic moment of the coil. (b) Find the maximum torque that the magnetic field can exert on the coil.
20 0 1.6 The Torque on a CurrentCarrying Coil Example The Torque Exerted on a CurrentCarrying Coil A coil of wire has an area of.0x104 m, consists of 100 loops or turns, and contains a current of A. The coil is placed in a uniform magnetic field of magnitude 0.15 T. (a) Determine the magnetic moment of the coil. (b) Find the maximum torque that the magnetic field can exert on the coil. (a) m magnetic moment ( )( )( 4 ) 4 NIA A.0 10 m A m (b) τ magnetic moment NIA Bsinφ ( A m )( 0.15 T) sin N m
21 1.6 The Torque on a CurrentCarrying Coil Application: The basic components of a dc motor. The brushes switches the direction of the current so that the torque is always in the same direction continuous rotation 1
22 1.8 Ampere s Law AMPERE S LAW FOR STATIC MAGNETIC FIELDS For any current geometry that produces a magnetic field that does not change in time, B µ oi net current passing through surface bounded by path positive sense for the current bounded is related to the direction of the path traveled by RHR Positive direction for bound currents
23 1.8 Ampere s Law Example An Infinitely Long, Straight, CurrentCarrying Wire Use Ampere s law to obtain the magnetic field. 3
24 1.8 Ampere s Law Example An Infinitely Long, Straight, CurrentCarrying Wire Use Ampere s law to obtain the magnetic field. 1. By symmetry, the magnetic field must have cylindrical symmetry: depends only on distance r to the wire.. The magnetic field should be generally perpendicular to the current (by Right Hand Rule). Choose a circular path (at constant r) to match the symmetry The magnetic field lines circulate around the line by RHR: traverse path in the CCW direction as seen from the top The field is parallel to the path and has constant magnitude on the path. B B µ I o ( ) µ I o B(π r) µ I µ o B I π r o 4
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