Faraday’s Law r The loop of wire shown below has a radius of 0.2 m, and is in a magnetic field that is increasing at a rate of 0.5 T/s. Since the area of the loop is constant, Δ(BA) = AΔB

I Have Proof! If area doesn’t change, but B does… Therefore, when A is unchanged, If area changed, but field remained constant, then you would end up with Δ(BA) = BΔA

Whiteboard Warmup! r The loop of wire shown below has a radius of 0.2 m, and is in a 0.8 T magnetic field. The loop is rotated by 90° in 2 seconds along the dotted line shown. The loop contains a 3 Ω resistor. a)What is the average induced emf? b) What is the induced current, and which way does it flow?

a)ε = 0.05 Volts b)I = ε/R = A counterclockwise Since B and A are constant

Reverse Rail Gun! A metal rod is pulled to the right at constant velocity, along two conducting rails shown below. What is the direction of the induced current through the circuit as a result?

Φ: Into page ΔΦ: Into page B induced : Out of page I induced : Counterclockwise Φ: Into page ΔΦ: Into page B induced : Out of page I induced : Counterclockwise

Whiteboard: What will be ε induced ?

!

Motional EMF When the amount of a loop’s area that in a magnetic field is changing at a constant rate, Faraday’s Law gives the result L is the side that is not changing. v is the rate of change of the side that is entering or leaving the field.

Using Motional EMF In terms of the quantities shown above, write an expression for a)The current through the resistor, and direction of current. b)The power output of the resistor in the circuit. c)The force required to pull the metal bar at constant velocity.

ε = BLv I = ε/R a) What is the current through the resistor? I = BLv/R Φ: Into page ΔΦ: Into page B induced : Out of page I induced : Counterclockwise Φ: Into page ΔΦ: Into page B induced : Out of page I induced : Counterclockwise I I I I

P = I 2 R b) What is the power output of the resistor? P = (BLv) 2 /R P = B 2 L 2 v 2 /R

c) What is the force required to pull the bar at constant speed? I Using RHR #2, you can determine that when there is a current flowing through the circuit, the moving metal bar will feel a magnetic force to the left. FBFB Therefore, to pull the bar at constant velocity, you must exactly balance out the magnetic force BIL.

c) What is the force required to pull the bar at constant speed? I FBFB F B = BIL F B = B(BLv/R)L F B = B 2 L 2 v/R In order to pull the bar at constant velocity, you must exactly match this force by pulling to the right

Let’s analyze the results for a second… P resistor = B 2 L 2 v 2 /R F = B 2 L 2 v/R P = Force x velocity! The power output of the resistor will be exactly equal to the power delivered to the system by pulling the rod.

Coils with Multiple Loops! Each coil acts as its own loop. If there are N coils, Just multiply by N!

Solenoid Φ = N * BA

Solenoids are useful! They multiply the magnetic flux, and therefore the induced emf, by the number of turns that the wire has

Ring Launcher! I

I B coil

This induces a current in the ring that opposes the field of the coil I B coil XSXS XSXS XSXS XSXS B ring I induced

I The current-carrying coil of wire acts like a magnet, with the field lines coming out of North and into South. N S

XSXS XSXS XSXS XSXS B ring I induced XSXS XSXS XSXS XSXS N S The current-carrying ring also acts like a magnet, with the field lines coming out of North and into South.

XSXS XSXS The net result looks like this! I B coil XSXS XSXS XSXS XSXS B ring I induced N S XSXS XSXS N S

Strong repulsion!!! I I induced N S N S

Whiteboard: Copper Tube Drop! N S v a) What will be the direction of the induced current in each of these sections of copper tube? b) Draw the “magnet” that each of these sections acts like. c) What will be the result when the magnet is dropped down the tube?

v Φ: Downward ΔΦ: Upward B induced : Downward Φ: Downward ΔΦ: Upward B induced : Downward I ind Φ: Downward ΔΦ: Downward B induced : Upward Φ: Downward ΔΦ: Downward B induced : Upward I ind

vv

v Attracted by the induced magnet above Repelled from the induced magnet below The magnet will fall slowly!!!