PHYS 155: Final Tutorial - University of...

PHYS 155:FINAL

Eric Peach

MagneticFields

The ThreeRight HandRules

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Force andTorque OnWire

Ampere’s Law

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Faraday’s Lawof Induction

Inductance

Summary

PHYS 155: Final Tutorial

Eric Peach

Saskatoon Engineering Students’ Society

[email protected]

April 13, 2015

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Overview

1 Magnetic Fields

2 The Three Right Hand Rules

3 Force and Torque On Wire

4 Ampere’s Law

5 Faraday’s Law of Induction

6 Inductance

7 Summary

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Tutorial Slides

These slides have been posted:

sess.usask.ca

homepage.usask.ca/esp991/

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Section 1

Magnetic Fields

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Magnetic Fields

Magnetic Fields have no start, no finish. There is NO point of origin.

∇ · B = 0

They are created by movement of charge.

∇× B = µ0J + µ0E0dE

dt

Changing fields can incite EMF in other objects.

∇× E = −dB

dt

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Magnetic Field Lines

Magnetic Field lines have no start, no finish.

Magnetic field strength B related to density of field lines.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Three Right Hand Rules

Very Spatial

Assume Positive Charge and Current

Find magnetic field direction in various cases

Find force on a moving charge or current carrying wire.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

RHR 1: Magnetic Field Around a Wire

Magnetic field circles around wires, in a direction CCW when wire is comingout of page.

Put your thumb along wire.

Your fingers curl around wire along magnetic field lines.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

RHR 2: Magnetic Field Inside a loop or Coil of Wire

Wire wound in a coil generates a magnetic field directed along the axis of thecoil.

Fingers wrap around the loop or coil, in the direction of the current.

Thumb points in the direction of magnetic field.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

RHR 2: Additional Consideration

We can find RHR No. 2 by applying Rule No. 1 to a coil instead of a straight wire!

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

RHR 3: Force on a Moving Charge

Charge or current moving in a magnetic field experiences a magnetic force.

Perpendicular to direction of movement

Perpendicular to magnetic field.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary


Thumb Along Direction of MovementIndex Finger along Magnetic FieldPalm / middle finger points to the force!Remember, this is for POSITIVE Charge. For negative charge likeelectrons, the force is opposite!

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary


PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Right Hand Rules: Summary

RHR #1: Straight Wire

Thumb Along Direction of Current

Fingers curl the way the magnetic field does.

RHR #2: Coil / Loop of Wire

Curl fingers around coil of wire in direction of current

Thumb tells you direction of magnetic field

RHR #3: Charge Moving in Magnetic Field

Thumb Along Movement of Positive Charge

Index finger along magnetic field direction

Palm or middle finger tells you direction of force on positive charge.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Some Helpful Magnetic Formulas

Magnetic Field for a long, straight wire

|B| =µ0I

2πr

Magnetic Field at Centre of Loop

|B| = Nµ0I

2R

Magnetic Field Inside Solenoid

|B| = µonI = µ0N

`I

Force on Moving Charge / Wire in Magnetic Field

F = q(v × B)

F = qvB sin θ

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Example 1 – Lorentz Force

A proton moving at 4.00× 106 m/s through a magnetic field of magnitude 1.70 Texperiences a magnetic force of magnitude 8.20× 10−13N. What is the anglebetween the proton’s velocity and the field?

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Example 1 – Solution

Draw a picture showing this effect. Lorentz force is

~F = q~v × ~B

so|~F | = qvB sin θ

Using magnitude of force, velocity and magnetic field, solve for θ.Answer: θ = 48.83 or 131.2.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Example 2 – Cyclotron Motion

A singly charged ion of mass m is accelerated from rest by a potential difference∆V . It is then deflected by a uniform magnetic field (perpendicular to the ion’svelocity) into a semicircle of radius R. Now a doubly charged ion of mass m′ isaccelerated through the same potential difference and deflected by the samemagnetic field into a semicircle of radius R ′ = 2R. What is the ratio of the massesof the ions?

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary


Write the kinetic energy of the particle.

KE = q∆V =1

2mv2 (1)

Write the conditition for centripetal acceleration

Fm = Fc , qvB =mv2

R(2)

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary


Sub (1) into (2) and solve for m in terms of q, B, ∆V and R.

m =B2

2∆VqR2

Solve for m′ in terms of q, B, ∆V and R. Note that all you have to do istweak your q, B & R values.

m =B2

2∆V(2q)(2R)2

Answer: m′ = 8.00×m.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Example 3 – Hold Straight Condition

A velocity selector consists of electric and magnetic fields described by theexpressions ~E = E k and ~B = B j, with B = 15.0mT. Find the value of E such thata 750-eV electron moving in the negative x direction is undeflected.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary


Draw your picture. Which way will the magnetic field try to direct theelectron? (Up)

Which way does the electric force need to apply? (Down). Which way shouldthe field be pointed? (Up)

Solve for electron velocity using KE = 12mv2

Draw FBD and write the condition for no deflection:

~FM + ~FE = 0

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary


Figure out the magnitude of the magnetic force on the electron.

~FM = q~v × ~B = 3.90× 10−14k N

Note that~FE = E k = − ~FM

Solve for E.

Answer: 244 kV/m

PHYS 155:FINAL

Eric Peach

MagneticFields


BREAK


Ampere’s Law

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Inductance

Summary

Break

See you in 10 Minutes

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Section 4

Force and Torque on Wiredue to Magnetic Field

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Force on a Wire

Extension of RHR #3.

F = I (~× ~B)

F = I `B sin θ

` is length of wire, whose direction is the direction of the current.

For 2 parallel wires, the force of attraction/repulsion is

|F| =µ02π

I1I2d

L

Attraction if current in same directionRepulsion if current in opposite direction

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Torque on a Loop

A loop of wire in a magnetic field expriences a torque, according to

~τ = NI [~A× ~B] τ = NIAB sin θ

the A vector has magnitude equal to area of loop, direction points in directionof magnetic field generated by loop (use RHR 2).

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Example 4 – Torque on Wire

A current of 17.0mA is maintained in a single circular loop of 2.00 mcircumference. A magnetic field of 0.800 T is directed parallel to the plane of theloop. What is the torque exerted on the loop by the magnetic field?

PHYS 155:FINAL

Eric Peach

MagneticFields


BREAK


Ampere’s Law

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Inductance

Summary


1 Draw a picture of the loop of wire and the field. Conceptualize: if currentflows in the loop, and the magnetic field is parallel to the plane of the loop,which side will try to lift up? Which side will be pushed down?

2 Solve for area of the loop using circumference.

3 Use Torque Formula to solve for torque.

~τ = NI [~A× ~B]

Answer: 4.33 ×10−3 N·m.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Example 5 – Attraction Between Wires

The current in the long, straight wire is I1 = 5.00 A and the wire lies in the planeof the rectangular loop, which carries a current I2 = 10.0 A. The dimensions in thefigure are c =0.100 m, a = 0.150 m, and ` = 0.450 m. Find the magnitude anddirection of the net force exerted on the loop by the magnetic field created by thewire.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary


1 Ask yourself: Which way will each part of the loop try to move?

2 Find the strength of the magnetic field due to the long wire at the left side ofthe loop.

B =µ0I12πc

3 Find the Lorentz force on the left side of the loop

~F = I2~× ~B

4 Find the magnetic field strength and the force on the right side of the loop.

B =µ0I1

2π(c + a)~F = I2~× ~B

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary


Note that the top and bottom parts of the loop experience forces too, butthey cancel out.

Subtract the force of the right side of the loop from the force of the left sideof the loop to find net force.

Answer: −2.7× 10−5i N

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Section 5

Ampere’s Law

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Ampere’s Law

Let’s us find the magnetic field around an enclosed loop of wire. Comes from

∇× B = µ0J + µ0E0dE

dt

This simplifies to∮~B · ~d` = µ0Ienc .

Pick some sort of symmetry, usually circular, around some source of current. Thenyou can find the magnetic field strength.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Example 6 – Ampere’s Law

A packed bundle of 100 long, straight, insulated wires forms a cylinder of radius R= 0.500 cm. If each wire carries 2.00 A, what are the (a) magnitude and (b)direction of the magnetic force per unit length acting on a wire located 0.200 cmfrom the centre of the bundle?

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary


1 Recall Ampere’s Law: ∮~B · ~d` = µ0Ienc

2 Figure out the current enclosed by a radius of 0.2cm. (It’s 32 amps).

3 If we integrate around a circle, centred at the centre of the wire, ~B will beconstant because of radial symmetry. So you get

B · (2πr) = µ0Ienc

4 Solve for ~B

5 Multiply by 2A, the current of 1 wire at that radius. (F/`=IB)

Answer: F/` = 6.4 mN.

PHYS 155:FINAL

Eric Peach

MagneticFields


BREAK


Ampere’s Law

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Inductance

Summary

Break

See you in 10 Minutes

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Section 7

Faraday’s Law of Inductionand Lenz’s Law

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Lenz’s Law

A loop or coil of wire that is exposed to achange in magnetic flux through the loop willinduce an EMF whose corresponding magneticfield will oppose the original change in flux.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Faraday’s Law of Induction

Mathematical Representation of Lenz’s Law

E = −N dΦ

dt= −N d

dt(AB cos θ)

So we get Back EMF for one of three reasons: Change in Area, Magnetic FieldStrength or Angle.Usually only one will change at a time in any given question.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Example 7 – Faraday Law of Induction

A coil formed by wrapping 50 turns of wire in the shape of a square is positioned ina magnetic field so that the normal to the plane of the coil makes an angle of 30.0

with the direction of the field. When the magnetic field is increased uniformly from200 µT to 600 µT in 0.400 s, an emf of magnitude 80.0 mV is induced in the coil.What is the total length of the wire in the coil?

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary


1 Recall Faraday Law of Induction

E = −N dΦ

dt= −N d

dt(AB cos θ)

2 Here, only the magnetic field is changing, so this simplifies to

E = −NAdB

dtcos θ

3 Solve for A, then find side length, then multiply by 4N to get length of wire.

Answer: 272m.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Bar and Rail Problems

1 Figure out the induced EMF

Lenz’s Law to find direction.Faraday’s Law for magnitude.

2 Using EMF, find current in loop (Ohm’s Law, KVL)

3 Using I and Lorentz force law, find magnetic force on bar.

4 Draw FBD of Bar, solve unknowns. Find any input force required.

5 Watch out for directions of B, v and F, especially on angled rails!

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Example 8 – Bar and Rail

The picture shows a bar of mass m = 0.200 kg that can slide without friction on a pair ofrails separated by a distance ` = 1.20 m and located on an inclined plane that makes anangle θ = 25.0 with respect to the horizontal. The resistance of the resistor is R = 1.00Ω and a uniform magnetic field of magnitude B = 0.500 T is directed downward,perpendicular to the ground, over the entire region through which the bar moves. Withwhat constant speed v does the bar slide along the rails?

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary


1 Solve for EMF using E = −`vB cos θ

2 Use Ohm’s Law to find current through loop. I = E/R3 Lorenz Force Law (F = qvB) to find the back EMF force on the bar (* Watch

out for direction here *).

4 Draw FBD to find relation between FM , N and Fg .

5 For constant velocity, you can relate FM and Fg . Solve for v

v =mg tan θR

`2B2 cos θ

Answer: 2.80 m/s

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Section 8

Inductance

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Inductance

Self Inductance

E = −N dΦ

dtEL = −LdI

dtL = µ0

N2

`A =

µ0µrN2A

`

Mutual Inductance

NsΦs = MIP Es = −M dIPdt

Energy In an Inductor

U =1

2LI 2

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Example 9 – Mutual Inductance

Two solenoids A and B, spaced close to each other and sharing the same cylindricalaxis, have 400 and 700 turns respectively. A current of 3.50 A in solenoid Aproduces an average flux of 300 µWb through each turn of A and a flux of 90.0µWb through each turn of B. (a) Calculate the mutual inductance of the twosolenoids. (b) What is the inductance of A? (c) What emf is induced in B whenthe current in A changes at a rate of 0.5 A/s ?

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary


1 Use the mutual inductance formula that has all the variables you know.

NSΦS = MIP

2 To find self inductance, use self inductance formula relating flux and current.

NΦ = LI

3 To find EMF, use EMF formula with mutual inductance.

ES = −M dIPdt

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary


It might be helpful to pay attention to the directions of the coils here. Rememberthat the ’-’ serves to remind you that it is back EMF, but the winding of the coilactually decides the direction of current.Answer: 0.018 H, 0.0343 H, -9mV.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Transformers

Just remeber the transformer turn ratio formula, relating current, voltage andnumber of turns in the Primary and Secondary coils.

NP

NS=

VP

VS=

ISIP

Use Ohm’s Law and P = VI to solve for everything else.

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary

Example 10 – Transformers

In the transformer shown, the load resistance RL is 50.0 Ω. The turns ratio N1/N2

is 2.50, and the rms source voltage is ∆Vs = 80.0 V. If a voltmeter across the loadresistance measures an rms voltage of 25.0 V, what is the source resistance Rs?

PHYS 155:FINAL

Eric Peach

MagneticFields


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Ampere’s Law

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Inductance

Summary


Remember the big important transformer equation.

NP

NS=

VP

VS=

ISIP

Using the secondary coil voltage and Ohm’s law, find current in secondary coil.

Using turns ratio, find the current in primary coil and voltage in primary acrossthe transformer.

KVL to find voltage across resistor, and Ohm’s Law to find resistance.

Answer: 87.5 Ω

PHYS 155:FINAL

Eric Peach

MagneticFields


BREAK


Ampere’s Law

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Inductance

Summary

Summary

1 Go through the slides (both mine and the ones on blackboard) and make up asolid formula sheet.

2 Have a good section on unit conversions, and what units you can expect fromcertain formulas.

3 Do the assignments. Each question has a recipe to solve it. Practise your LenzLaw to anticipate direction of magnetic fields and back EMF.

4 Watch out for weird units on the exam.

PHYS 155:FINAL

Eric Peach

MagneticFields


BREAK


Ampere’s Law

BREAK


Inductance

Summary

Good Luck!

These slides have been posted:

sess.usask.ca

homepage.usask.ca/esp991/

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