Physics 4. Magnetic Forces and Fields. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Transcription

1 Physics 4 Magnetic Forces and Fields

2 What creates a magnetic field? Answer: MOVING CHARGES What is affected by a magnetic field? Answer: MOVING CHARGES We have a formula for magnetic force on a moving charge: F mag q v B q v B sin( )

3 What creates a magnetic field? Answer: MOVING CHARGES What is affected by a magnetic field? Answer: MOVING CHARGES We have a formula for magnetic force on a moving charge: F mag q v B q v B sin( ) This is a vector cross-product. We need a right-hand-rule to find the direction of this force. How to find the direction: 0)Use your RIGHT HAND 1)Fingers start in the direction of the charge s velocity. 2)Curl fingers toward the direction of the magnetic field. 3)Thumb points in the direction of the magnetic force on the charge. 4)If the charge is negative, flip your hand over. Important notes: If velocity is aligned with the magnetic field, force is zero. Magnetic force is always perpendicular to both the velocity Prepared and the by Vince B-field. Zaccone

4 A uniform magnetic field is directed into the page. Find the direction of the magnetic force on charge A, a positive charge. B v A A

5 A uniform magnetic field is directed into the page. Find the direction of the magnetic force on charge A, a positive charge. B A F mag v A -Fingers start along A s velocity (up in the picture) -Fingers bend toward B-field (into the page) -Thumb sticks out to the left.

6 A uniform magnetic field is directed into the page. Find the direction of the magnetic force on charge B, a positive charge. B v X X B X X X X B

7 A uniform magnetic field is directed into the page. Find the direction of the magnetic force on charge B, a positive charge. B v X X B X X X X B -Fingers start along B s velocity (up and to the left in the picture) -Fingers bend toward B-field (into the page) -Thumb sticks out down and to the left. F mag

8 A uniform magnetic field is directed into the page. Find the direction of the magnetic force on charge C, a negative charge. B C v C

9 A uniform magnetic field is directed into the page. Find the direction of the magnetic force on charge C, a negative charge. B C v C -Fingers start along C s velocity (to the right in the picture) -Fingers bend toward B-field (into the page) -Thumb sticks out upward. -Force is downward. (negative charges are pushed in the opposite direction) F mag

10 A uniform magnetic field is directed into the page. Now that we can find the force on a charge, we should be able to predict its trajectory. Notice that the force is always perpendicular to the velocity. This will yield a circular path. In other words, the magnetic force is a centripetal force. B v A A F mag F mag C v C Notice that in this diagram, A is a positive charge, and moves counter-clockwise. Since B is a negative charge, it moves the opposite direction - clockwise. So you can tell the sign of the charge by which direction it rotates in a magnetic field. What path would a neutral charge follow?

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12 Application: Mass Spectrometer Ions are fired into a region of constant magnetic field. Magnetic force pushes it into a circular path and the impact location will determine the mass/charge ratio of the ion. F mag = F cent qvb = mv2 r r = mv qb Velocity selector

13 Application: Mass Spectrometer Here is a problem from the book:

14 Application: Mass Spectrometer Here is a problem from the book: We can use the radius formula we just derived. How do we get the potential difference involved?

15 Application: Mass Spectrometer Here is a problem from the book: We can use the radius formula we just derived. How do we get the potential difference involved? Recall the formula for electrical potential energy: U elec =qv Since they are accelerated from rest, this electrical potential energy becomes the kinetic energy of the particle.

16 Application: Mass Spectrometer Here is a problem from the book: We can use the radius formula we just derived. How do we get the potential difference involved? Recall the formula for electrical potential energy: U elec =qv Since they are accelerated from rest, this electrical potential energy becomes the kinetic energy of the particle. The charge on each particle is just e, the elementary charge. So we can write a formula: 1 2 mv2 = ev

17 Application: Mass Spectrometer Here is a problem from the book: We can use the radius formula we just derived. How do we get the potential difference involved? Recall the formula for electrical potential energy: U elec =qv Since they are accelerated from rest, this electrical potential energy becomes the kinetic energy of the particle. The charge on each particle is just e, the elementary charge. So we can write a formula: 1 2 mv2 = ev Rearrange to solve for velocity: v = 2eV m And plug back into our radius formula: r = m 2eV m eb = 2eVm eb Subtracting the radius for the two isotopes gives their desired result.

18 Application: Mass Spectrometer Here is a problem from the book: We can use the radius formula we just derived. How do we get the potential difference involved? Recall the formula for electrical potential energy: U elec =qv Since they are accelerated from rest, this electrical potential energy becomes the kinetic energy of the particle. The charge on each particle is just e, the elementary charge. So we can write a formula: 1 2 mv2 = ev Rearrange to solve for velocity: v = 2eV m And plug back into our radius formula: r = m 2eV m eb = 2eVm eb Subtracting the radius for the two isotopes gives their desired result. Part b) is just plugging in numbers. Answer: V = 3317 Volts

19 A uniform magnetic field is directed into the page. Find the direction of the magnetic force on the wire, with current flowing as shown. B I

20 A uniform magnetic field is directed into the page. Find the direction of the magnetic force on the wire, with current flowing as shown. B F mag I Right-Hand Rule (same as before) Pretend that positive charges are flowing in the direction of the current. -Fingers start along A s current (up in the picture) -Fingers bend toward B-field (into the page) -Thumb sticks out to the left. Here is a formula for the force on a wire in a B-field: F = I l B = IlB sin θ

21 A uniform magnetic field is directed into the page. Find the direction of the magnetic force on the wire, with current flowing as shown. B F mag I Right-Hand Rule (same as before) Pretend that positive charges are flowing in the direction of the current. -Fingers start along A s current (up in the picture) -Fingers bend toward B-field (into the page) -Thumb sticks out to the left. Here is a formula for the force on a wire in a B-field: F = I l B = IlB sin θ If your wire is not straight you can do an integral along the wire instead: d F = Id l B = IlB sin θ

22 An example from the book. Try it (it s a fun one I promise!)

23 An example from the book. Try it (it s a fun one I promise!) The long way to do this problem involves integrating along the wire. This involves 3 sections 2 straight parts and one semicircular arc.

24 An example from the book. Try it (it s a fun one I promise!) The long way to do this problem involves integrating along the wire. This involves 3 sections 2 straight parts and one semicircular arc. The shortcut uses symmetry to our advantage. The force on each section of wire is perpendicular to the wire. If we break the force into components, the vertical components cancel, leaving just the horizontal components. It comes out the same as a wire that is straight instead of curved. This saves us the trouble of integrating and we can just multiply it out. F = IlB sin θ = 3.40A 3.00m 2.20T sin 90 = 22.4N

25 Application: Torque on a current loop This is the basis for electric motors.

26 Application: Torque on a current loop Here is a problem from the book:

27 Application: Torque on a current loop Here is a problem from the book: The net force will be zero in both parts. Each side of the loop feels a magnetic force, and they cancel.

28 Application: Torque on a current loop Here is a problem from the book: The net force will be zero in both parts. Each side of the loop feels a magnetic force, and they cancel. As for the torque, in part a) we can see from the diagram that the coil is in an unstable equilibrium position. The field lines go straight through the loop, but the direction of the current makes the dipole moment of the coil point out of the page (opposite the field). In the book s explanation, the angle φ=180. So the torque is zero. If the current in the wire went the opposite way, the coil would be in a stable equilibrium.

29 Application: Torque on a current loop Here is a problem from the book: The net force will be zero in both parts. Each side of the loop feels a magnetic force, and they cancel. As for the torque, in part a) we can see from the diagram that the coil is in an unstable equilibrium position. The field lines go straight through the loop, but the direction of the current makes the dipole moment of the coil point out of the page (opposite the field). In the book s explanation, the angle φ=180. So the torque is zero. If the current in the wire went the opposite way, the coil would be in a stable equilibrium. For part b) we will have to do some calculating. Tilting the loop by 30 makes the angle φ=150. Using the book s formula, we get: τ = IAB sin φ = 1.40A 0.22m 0.35m 1.50T sin 150 = 0.08 N m

30 Application: Torque on a current loop Here is a problem from the book: The net force will be zero in both parts. Each side of the loop feels a magnetic force, and they cancel. As for the torque, in part a) we can see from the diagram that the coil is in an unstable equilibrium position. The field lines go straight through the loop, but the direction of the current makes the dipole moment of the coil point out of the page (opposite the field). In the book s explanation, the angle φ=180. So the torque is zero. If the current in the wire went the opposite way, the coil would be in a stable equilibrium. For part b) we will have to do some calculating. Tilting the loop by 30 makes the angle φ=150. Using the book s formula, we get: τ = IAB sin φ = 1.40A 0.22m 0.35m 1.50T sin 150 = 0.08 N m This is the magnitude. The direction of the torque is along the axis (up in their diagram). The torque will always be oriented in such a way as to rotate the coil so that its dipole moment aligns with the field. In this problem that means the coil needs to flip all the way over.

31 Application: Torque on a current loop Here is a problem from the book: The net force will be zero in both parts. Each side of the loop feels a magnetic force, and they cancel. As for the torque, in part a) we can see from the diagram that the coil is in an unstable equilibrium position. The field lines go straight through the loop, but the direction of the current makes the dipole moment of the coil point out of the page (opposite the field). In the book s explanation, the angle φ=180. So the torque is zero. If the current in the wire went the opposite way, the coil would be in a stable equilibrium. For part b) we will have to do some calculating. Tilting the loop by 30 makes the angle φ=150. Using the book s formula, we get: τ = IAB sin φ = 1.40A 0.22m 0.35m 1.50T sin 150 = 0.08 N m This is the magnitude. The direction of the torque is along the axis (up in their diagram). The torque will always be oriented in such a way as to rotate the coil so that its dipole moment aligns with the field. In this problem that means the coil needs to flip all the way over. You can probably imagine what would happen if the loop was allowed to rotate it would flip over, overshoot, then flip back and forth, oscillating about the stable equilibrium point. This is *almost* an electric motor. What needs to be done to make it keep rotating is to reverse the current every time it flips so that the torque always points the same way.