Forces & Magnetic Dipoles. Phys 122 Lecture 18 D. Hertzog

Transcription

1 orces & Magnetic Dipoles µ = τ U AI = µ = µ θ θ. Phys 122 Lecture 18 D. Hertzog µ

2 usiness Regrade requests by 4 pm riday (no eceptions) Solutions/Key posted on home page

3 Last Time: The Lorentz orce and the RHR = qv 1. Point ingers in direction of v 2. Curl all fingers in direction of qv z 3. Thumb points in direction of And a few challenging Clickers: y Only z since it does not change the speed 1 = 2 ;; both are perpendicular to Work done by magnetic field is = 0;; is always perpendicular to d

4 Magnetic orce on a Current What is the total force d on a length dl of current-carrying wire in field? Current described by: n charges q per volume Each has velocity= v Wire cross-section = A. orce on a single charge = orce on charges moving in line segment dl = Recall Current = qv d = q tot ( v ) = q( na)( dl)( v ) dq I = * = qnav Þ dt d = Idl N S or a straight wire of length L carrying a current I, the force on it is therefore : = " IL

5 *Reminder from Prelecture What units result when a velocity v is multiplied by an area A? Volume per second A simple and useful way to picture this is to think of water flowing in a garden hose. If the speed of the water is known (v = 10 m/s = 1,000 cm/s for eample) and the cross-sectional area of the hose is also known (A = 10 cm 2 for eample), then the volume of water that travels past any point of the hose is every second is v*a = 10,000 cm 3 /s = 10 liters/s. In other words, if we know the velocity of the water and the area of the hose we can easily figure out the volume of water transported by the hose per unit time. If we multiply this by the density of water (i.e., 10 liters/s * 1 kg/liter = 10 kg/s) then we immediately discover how much mass is moving past a point in the hose the hose per unit time. This has an eact analogy with current in a wire. Think of the wire as the hose and the current flowing in the wire as the water. The average velocity of the charges v avg times the area of the wire A tells us the volume of the charge carriers that moves through the wire per unit time. If we multiply this by the number density of the charge carriers n we find the number of charge carriers per unit time flowing in the wire, and if we multiply this by the charge carried by each one q we find the charge per unit time flowing in the wire. In other words, the current in the wire I is equal to the product qnav avg.

6 Work and Magnetic ields Magnetic fields do no work on free charges Magnetic fields can do work on current carrying wires

7 Shout-out Clicker A square loop of wire is carrying current in the counterclockwise direction. A horizontal uniform magnetic field points to the right = IL A C

8 Shout-out Clicker A square loop of wire is carrying current in the counterclockwise direction. A horizontal uniform magnetic field points to the right = IL A C

9 Shout-out Clicker A square loop of wire is carrying current in the counterclockwise direction. A horizontal uniform magnetic field points to the right = IL What is the force on section d- a of the loop? A) Zero ) Out of the page C) Into the page

10 Shout out CheckPoint 2 A square loop of wire is carrying current in the counterclockwise direction. A horizontal uniform magnetic field points to the right A C

11 (real) Clicker A current I flows in a wire which is formed in the shape of an isosceles triangle as shown. A constant magnetic field eists in the -z direction. What is y, net force on the wire in the y-direction? (a) y < 0 (b) y = 0 (c) y > 0 The forces on each segment are determined by: y = IL I I I 1 θ θ 2 θ Lsinθ θ rom symmetry, = 0 or the y-component: Therefore: 1 y = 2 y = ILsinθ y 3 = 2ILsinθ = + + = y 1y 2y 3y 0 3

12 Clicker of a CheckPoint 4 A square loop of wire is carrying current in the counterclockwise direction. A horizontal uniform magnetic field points to the right y In which direction will the loop rotate? (assume the z ais is out of the page) A) Around the ais ) Around the y ais C) Around the z ais D) It will not rotate 50%

13 Torque on a loop Coil has width w (the side you see) and length L (into the screen). Torque is given by: Þ τ r w τ = 2 sinθ 2 θ w. = IL Þ τ = IA sinθ where A = wl = area of loop r maimum τ occurs when the plane of the loop is parallel to, which is when θ = 90 r

14 Some pictures

15 CheckPoint 6 A square loop of wire is carrying current in the counterclockwise direction. A horizontal uniform magnetic field points to the right R τ = R 42%

16 Magnetic Dipole Moment Define magnetic dipole moment of a current loop as: magnitude: µ = AI direction: to plane of the loop in the direction the thumb of right hand points if fingers curl in direction of current. Torque on loop is then: τ = AI sinθ Note: if loop consists of N turns, µ = NAI Þ τ = µ θ µ θ. Remember this: The torque always wants to line µ up with

17 Do we know the directions?

18 Clicker A circular loop of radius R carries current I. A constant magnetic field eists in the + direction. Initially the loop is in the -y plane. How will the coil rotate? y R I y y (a) R (b) (c) It will not rotate R I I The coil will rotate if the torque on it is non-zero: τ = µ The magnetic moment µ is in +z direction. Therefore the torque τ is in the +y direction. Therefore the loop will rotate as shown in (b).

19 Potential Energy of Dipole Work required to change the orientation of a magnetic dipole in the presence of a magnetic field. Define potential energy U (with zero at position of ma torque) corresponding to this work. θ µ θ. U τdθ Þ U = µ

20 CheckPoint 8 τ = µ iggest when µ good

21 Magnetic ield can do Work on Current W = τdθ r r r τ = µ = µ sin( θ) sign in integration because direction of rotation is decreases θ W θ = µ sin( θ ) dθ = µ cos( θ ) = µ 90 ΔU = W 1.5 Defined U = 0 at position of maimum torque U µ U µ

22 CheckPoint 10 θ U = +µcosθ U = 0 θ U = µcosθ U = µ U

23 CheckPoint 12 In order to rotate a horizontal magnetic dipole to the three postions shown, which one requires the most work done by the magnetic field? θ a φ a θ c Y YOU C): ): A): Y IELD W W W W by _ field Y WHOM? = ΔU = U U = µ i U = µ ( µ cosθ ) = µ (1 cosθ ) by _ field c by _ field = µ 0 = µ by _ field = µ ( µ cosθ ) = µ (1 + cosφ ) a f c a

24 (similar) Potential Energy of Dipole (this is an important slide) µ µ µ = 0 = µ X τ = 0 θ = 0 θ = 90 θ = 180 U = -µ U = 0 U = µ (minimum) (maimum)