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a g b From these observations alone,

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Presentation on theme: "a g b From these observations alone,"— Presentation transcript:

1 a g b From these observations alone,
B-field into screen a g b From these observations alone, what definite conclusions can be made? as are positively charged, bs negative. as are negatively charged, bs positive. can only say a,b oppositely charged.

2 a g b From these observations alone,
B-field into screen a g b From these observations alone, how can we compare the magnitude of their charges?  qa <qb 4. charges can’t be  qa =qb directly compared  qa >qb

3 a g b From these observations alone,
B-field into screen a g b From these observations alone, what definite conclusions can be made? as are more massive than bs. as are more massive than gs. bs are more massive than as. The masses cannot be directly compared.

4 What produces a gravitational field? Mass
A gravitational field exerts a force on? Mass What produces an electric field? Electric charge An electric field exerts a force on? What produces a magnetic field? Moving electric charge A magnetic field exerts a force on? Moving electric charge?

5 Direction of Magnetic Force
out of page in to page tail Drawing vectors in 3D head Direction of magnetic force is “sideways” force is perpendicular to both v and B use “right-hand rule” to find direction F = q v B sinq

6 N A positive charge enters a magnetic field region as shown. The B-field accelerates it in what direction? 1) up 2) down 3) right 4) left 5) into the page 6) out of the page + S

7 ConcepTest Magnetic Force
A positively charged beam enters into a magnetic field region as shown. What is the direction of B? 1) + y (up) 2) – y (down) 3) + x (right) 4) + z (out of page) 5) – z (into page) x y

8 Trajectory in a Constant Magnetic Field
Suppose a charge q enters a B field with velocity v What will be the path q follows? Force is always ^ to velocity and B. What is the path? x x x x x x x x x x x x x x B v F +q R Path will be a circle with some radius of curvature, R.

9 Radius of Circular Orbit
magnetic force: x x x x x x x x x x x x x x v B +q R F centripetal accel: Newton's 2nd Law: Þ Þ This has useful experimental consequences !

10 ConcepTest Magnetic Force
x x x x x x x x x x x x Two particles of the same charge enter a magnetic field with the same speed. Which one has the bigger mass? A B 1) A 2) B 3) both masses are equal 4) impossible to tell without weighing the particles

11 as are ionized Helium (bare Helium nuclei)
g b as are ionized Helium (bare Helium nuclei) 2-protons, 2-neutrons (positively charged) bs are simply electrons(negatively charged) qa = -2qb ma=7296mb

12            V=Ed v=? V

13            V=Ed R=? v= /2Vm/q /B V

14 Sum of the forces on the particle?
Velocity Selector Consider a positively charged ion entering a region where the electric and magnetic fields are uniform and perpendicular to each other. If the particle moves in a straight line, what is its velocity in terms of E and B? B x x x x x E For the magnetic force: direction magnitude up F = qvB For the electric force? direction magnitude down F = qE Sum of the forces on the particle? Zero (not accelerating)  |FE| = |FB|  qE = qvB v = E / B

15 Ratio of charge to mass for an electron
e– “gun” An electron is accelerated from rest across a potential difference and then enters a region of uniform magnetic field, as shown at right. What is the “charge to mass ratio”, q/m, of the electron? DV e– x x x x R What is the speed of the electron? ½ mv2 = qV (Work-Energy Theorem) What is the radius of the electron’s orbit? R = mv / qB (Earlier today) Algebra: determine q/m (Solve second Eq for v and plug into first) q / m = 2V / R2B2 B

16 ConcepTest I 2 3 1 I I I If all wires carry the same current I, for which of the loops above is the magnitude of the net force greatest? A) Loop 1 B) Loop 2 C) Loop 3 D) same for all

17 ConcepTest Magnetic Force
A rectangular current loop is in a uniform magnetic field. What direction is the net force on the loop? (a) + x (b) + y (c) zero (d) – x (e) – y x z y B

18 ConcepTest If there is a DC current in the loop in the direction shown, the loop will A) move up B) move down C) rotate clockwise D) rotate counterclockwise E) some combination of moving and rotating N S

19 N s a b Current, I, flows as shown through a rectangular loop
ab immersed in a uniform magnetic field B. a b N s

20 N s a b The force on the top segment of the rectangular loop is 1)up.
2)down. 3)into screen. 4)out. 5)left. 6)right 7)zero. N s

21 N s a b The force on the bottom segments of the rectangular loop is
1)up. 2)down. 3)into screen. 4)out. 5)left. 6)right 7)zero. N s

22 N s a b The force on the left segment of the rectangular loop is 1)up.
2)down. 3)into screen. 4)out. 5)left. 6)right. 7)zero. N s

23 N s a b The force on the right segment of the rectangular loop is
1)up. 2)down. 3)into screen. 4)out. 5)left. 6)right. 7)zero. N s

24 Viewed from above, looking down
I q

25

26 1. as are positively charged, bs negative.
ANSWERS to CONCEPT QUESTIONS 1. as are positively charged, bs negative. 4. charges can’t be directly compared R depends not only on the size of the charge, \ but also the mass & velocity of the moving charge! 4. The masses cannot be directly compared. See above! 4) left + charge = Right hand. Fingers down (direction of B-field) Thumb into page (direction of moving charge). Palm points left (direction of force on the charge). 5) – z (into page) 1) A R=mv/qB : the larger the mass, M, the greater R must be (all other varibles being constant).

27 ANSWERS to CONCEPT QUESTIONS
A) Loop 1 Everywhere beneath the wire the magnetic field points into the page. The top segments are therefore pushed down, the bottom segments up. But not equally! Near the wire the magnetic field is stronger, so the push is stronger. (c) zero C) rotate clockwise 2)down. Current is at an angle but still crosses field lines… For the top loop crossing lines as the current moves into the page, for the bottom, as it comes out of the page. 1)up. 3)into screen. Current runs up through the left (front) segment, down through the right (back) segment. It crosses field lines in both cases. Unlike the pushes up and down on the top/bottom segments, these do not cancel…they are off-centered pushes that produce a torque! 4)out.


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