1norah ali al- moneef example : sketch the path and calculate the acceleration. e=2000 n/c, and v =...

1Norah Ali Al- moneef

Example: Sketch the path and calculate the acceleration. E=2000 N/C, and v = 2,000 m/s (figure on the right)


Example

A wire carries a current of 25 A. What is the magnetic field 10 cm from this wire?

B = oI / 2r B = (4 X 10-7 T m/A)(25A) (2)(0.10 m)B = 5.0 X 10-5 T


i1

i2

i3

ds

i4

)( 3210loop

iiisdB

Thumb rule for sign; ignore i4

Ampere’s law: Closed Ampere’s law: Closed Loops Loops


Example

A 10 cm long solenoid has a total of 400 turns of wire and carries a current of 2.0 A. Calculate the magnetic field inside the solenoid.

n = 400 turns/0.10 m = 4000 m-1

B = onI

B = (4 X 10-7 T m/A)(4000m-1)(2.0 A)B = 0.01 T


x x x x x x

x x x x x x

x x x x x x

x x x x x x

x x x x x x

B

P

x x x x x x

x x x x x x

x x x x x x

x x x x x x

x x x x x x

B

P v

Sample Problem: Sketch the path and calculate the acceleration. B = 2.0 T, and v = 2,000 m/s (for figure on the right).


Magnetic Force

• What can be concluded about the sign of the charge of each particle?

1

2

3

positive

neutral

negative


ExampleAn electron travels at 2.0 X 107 m/s in a plane perpendicular to a

0.010-T magnetic field. Describe its path.

Path is circular (right-hand rule, palm positive)F = m v2/rqvB = mv2/rr = mv /qB

r = mv / qBr = (9.1 X 10-31 kg)(2.0 X 107 m/s )

(1.6 X 10-19 C)(0.010 T)r = 0.011 m


• Sample Problem: A wire carries a current of 2.40 Amperes through a uniform magnetic field B=1.6 k Tesla. What is the force on a 0.75 meter long section of the wire if the current moves in the +x direction?


• Sample problem: How much current must be flowing through a horizontal 20.0 meter long wire to suspended in a 5.0 T magnetic field if the wire has a mass of 0.20 kg?

• If the current is flowing north, in what direction must be the magnetic field?


Example

What is the force on a wire carrying 30 A through a length of 12 cm? The magnetic field is 0.90 T and the angle is 60o.

F = IlBsinF = (30A)(0.12 m)(0.90 T)sin60o

F = 2.8 N into the page


ExampleTwo parallel wires have currents of I1 and I2 in exactly the same direction. Define the magnitude and direction of the force they exert on each other.


ExampleA loop of wire carries 0.245 A and is placed in a magnetic

field. The loop is 10.0 cm wide and experiences a force of 3.48 X 10-2 N downward (on top of gravity). What is the strength of the magnetic field?

F = IlBsinF = IlBsin90o

F = IlB(1) F = IlBB = F

IlB = 3.48 X 10-2 N = 1.42 T

(0.245 A)(0.100m)


15

Example• A magnetic pole face has a rectangular section

having dimensions 200 mm by 100 mm. If the total flux emerging from the pole is 150 μ Wb. Calculate the B.

Solution:• Φ = 150 μ Wb = 150 x 10-6 Wb• c.s.a = 200 x 100 = 20000 mm2 = 20000 x 10-6 m2

B = Φ/A = 150 x 10-6/20000 x 10-6 = 0.0075 T or 7.5mT

Norah Ali Al- moneef

Force Between Two Parallel Currents

• Force on I2 from I1

– RHR Force towards I1


– RHR Force towards I2

• Magnetic forces attract two parallel currents

I1I2

0 1 0 1 22 2 1 2 2 2

I I IF I B L I L L

r r

I1I2

0 2 0 1 21 1 2 1 2 2

I I IF I B L I L L

r r


Force Between Two Anti-Parallel Currents


– RHR Force away from I1


– RHR Force away from I2

• Magnetic forces repel two antiparallel currents

I1I2I1I2

0 1 0 1 22 2 1 2 2 2

I I IF I B L I L L

r r

0 2 0 1 21 1 2 1 2 2

I I IF I B L I L L

r r


Parallel Currents (cont.)

• Look at them edge on to see B fields more clearly

Antiparallel: repel

FF

Parallel: attract

F F

B

BB

B

2 1

2

2

2

1

11


ExampleTwo wires in a 2.0 m long cord are 3.0 mm apart.

If they carry a dc current of 8.0 A, calculate the force between the wires.

F = o I1I2 l

2 LF = (4 X 10-7 T m/A)(8.0A)(8.0A)(2.0m)

(2(3.0 X 10-3 m)F = 8.5 X 10-3 N


ExampleThe top wire carries a current of 80 A. How much

current must the lower wire carry in order to leviate if it is 20 cm below the first and has a mass of 0.12 g/m?F = o l I1I2

2 Lmg = o l I1I2

2 L

Solve for I2

I2 = 15 A20Norah Ali Al- moneef

Example: Use Ampere’s Law to derive the magnetic field a distance r from the center of a wire of radius R carrying current Io, where r<R.

Io

r

B

R21Norah Ali Al- moneef

• Formula found from Ampere’s law– i = current– n = turns / meter

– B ~ constant inside solenoid– B ~ zero outside solenoid– Most accurate when L>>R

• Example: i = 100A, n = 10 turns/cmn = 1000 turns / m

0B in

7 34 10 100 10 0.13TB


Sample problem: Using Ampere’s Law, show that the magnetic field inside a solenoid has magnitude B = oIon


example: Use Ampere’s Law to derive the magnetic field at various spots around the Toroid.


25

Example• Find the magnetic field in which the flux in

0.1m2 is 800μWb.

• Solution B = Φ/A = 800μWb/0.1m2 = 8000μT


Example1. In an area of 2 m2 there are 6 field lines each representing a flux

of 1 Wb. What is the flux density?

B = Φ / A = (6 x 1) / 2 = 3 T

2. A magnetic field has a magnitude of 0.0078 T and is uniform over a circular surface that has radius of 0.10 m. The field is orientated at an angle of of θ= 25o with respect to the surface. What is the flux through the surface?

Φ = BA cos θ= 0.0078 x (π x 0.12) cos 75 = 1 x 10-4 Wb


Example

In a coil of 3 m2 and 4 turns there are 12 field lines each representing a flux of 2 Wb. What is the flux density?

B = Φ / AN = (12 X 2) / (3 X 4) = 2 T


• We have observed that force is exerted on a charge by either and E field or a B field (when charge is moving):

• Consequences of the Lorentz Force:– A B field can exert a force on an electric current (moving charge)– A changing B-field (such as a moving magnet) will exert a magnetic force on

a static charge, producing an electric current → this is called electromagnetic induction

• Faraday’s contribution to this observation:– For a closed loop, a current is induced when:1. The B-field through the loop changes2. The area (A) of the loop changes3. The orientation of B and A changes

on a charge {together this is the Lorentz Force}F = qE + qv B

q

v

N

S

B

F

q

v

N

S

B

F

• A current is induced ONLY when any or all of the above are changing

• The magnitude of the induced current depends on the rate of change of 1-3

Moving charge

Moving magnet28

Magnetic Flux• Faraday referred to changes in B field, area and orientation as

changes in magnetic flux inside the closed loop• The formal definition of magnetic flux (B(analogous to

electric flux)

When B is uniform over A, this becomes:

• Magnetic flux is a measure of the # of B field lines within a closed area (or in this case a loop or coil of wire)

• Changes in B, A and/or change the magnetic fluxFaraday’s Law: changing magnetic flux induces electromotive

force (& thus current) in a closed wire loop

B = B dA

A

B

B = BA cos


In a series of experiments Michael Faraday in England

and Joseph Henry in the US were able to generate

electric currents without the use of batteries

Below we describe some of the

Faraday's experiments

se experiments that

helped formulate whats is known as "Faraday's law

of induction"

The circuit shown in the figure consists of a wire loop connected to a sensitive

ammeter (known as a "galvanometer"). If we approach the loop with a permanent

magnet we see a current being registered by the galvanometer. The results can be

summarized as follows:

A current appears only if there is relative motion between the magnet and the loop

Faster motion results in a larger current

If we

1.

2.

3. reverse the direction of motion or the polarity of the magnet, the current

reverses sign and flows in the opposite direction.

The current generated is known as " "; the emf that appears induced current

is known as " "; the whole effect is called " "induced emf induction 30Norah Ali Al- moneef

In the figure we show a second type of experiment

in which current is induced in loop 2 when the

switch S in loop 1 is either closed or opened. When

the current in loop 1 is constant no induced current

is observed in loop 2. The conclusion is that the

magnetic field in an induction experiment can be

generated either by a permanent magnet or by an

electric current in a coil.

loop 1loop 2

Faraday summarized the results of his experiments in what is known as

" "Faraday's law of induction

An emf is induced in a loop when the number of magnetic field lines that

pass through the loop is changing


N S

magnet motion

Bar magnet approaches the loop

with the north pole facing the loop.

2. Opposition to flux change

Example a :

B

As the bar magnet approaches the loop the magnet field points towards the left

and its magnitude increases with time at the location of the loop. Thus the magnitude

of the loop magnetic flux also

B

increases. The induced current flows in the

(CCW) direction so that the induced magnetic field opposes

the magnet field . The net field . The induced current is t

i

net i

B

B B B B

counterclockwise

B B

hus trying

to from increasing. Remember it was the increase in that generated

the induced current in the first place.

prevent


N S

magnet motion

Bar magnet moves away from the loop

with north pole facing the loop.


Example b :

B

As the bar magnet moves away from the loop the magnet field points towards the left

and its magnitude decreases with time at the location of the loop. Thus the magnitude

of the loop magnetic flux

B

also decreases. The induced current flows in the

(CW) direction so that the induced magnetic field adds to

the magnet field . The net field . The induced current is thus

i

net i

B

B B B B

clockwise

B B

trying

to from decreasing. Remember it was the decrease in that generated


prevent


S N

magnet motion

Bar magnet approaches the loop

with south pole facing the loop.


Example c :

B

As the bar magnet approaches the loop the magnet field points towards the right

and its magnitude increases with time at the location of the loop. Thus the magnitude

of the loop magnetic flux als

B

o increases. The induced current flows in the

(CW) direction so that the induced magnetic field opposes

the magnet field . The net field . The induced current is thus try

i

net i

B

B B B B

clockwise

B B

ing

to from increasing. Remember it was the increase in that generated


prevent


S N

magnet motion

Bar magnet moves away from the loop

with south pole facing the loop.


Example d :

As the bar magnet moves away from the loop the magnet field points towards the

right and its magnitude decreases with time at the location of the loop. Thus

the magnitude of the loop magnetic flux

B

B also decreases. The induced current

flows in the (CCW) direction so that the induced magnetic field

adds to the magnet field . The net field . The induced cur

i

net i

B

B B B B

counterclockwise

B B

rent is thus

trying to from decreasing. Remember it was the decrease in that

generated the induced current in the first place.

prevent


Induced Current• The next part of the story is that a changing

magnetic field produces an electric current in a loop surrounding the field– called electromagnetic induction, or Faraday’s Law


• (a) A bar magnet is moved to the right toward a stationary loop of wire. As the magnet moves, the magnetic flux increases with time

• (b) The induced current produces a flux to the left to counteract the increasing external flux to the right


N S

Faraday’s Experiment - 1:

G

NS

G G

G

NS N S

NSN S


Observe:

i) the relative motion between the coil and the magnet

ii) the induced polarities of magnetism in the coil

iii) the direction of current through the galvanometer and hence the

deflection in the galvanometer

iv) that the induced current (e.m.f) is available only as long as there is

relative motion between the coil and the magnet

Note: i) coil can be moved by fixing the magnet

ii) both the coil and magnet can be moved (towards each other or

away from each other) i.e. there must be a relative velocity between

them

iii) magnetic flux linked with the coil changes relative to the positions

of the coil and the magnet

iv) current and hence the deflection is large if the relative velocity

between the coil and the magnet and hence the rate of change of

flux across the coil is more 41Norah Ali Al- moneef

E

N S NS

Faraday’s Experiment - 2:

N S

K

N S

When the primary circuit is closed current grows from zero to maximum value.

During this period, changing current induces changing magnetic flux across the primary coil.

This changing magnetic flux is linked across the secondary coil and induces e.m.f (current) in the secondary coil.

Induced e.m.f (current) and hence deflection in galvanometer lasts only as long as the current in the primary coil and hence the magnetic flux in the secondary coil change.

P S

S

K G

P

E G


When the primary circuit is open current decreases from maximum value to zero.

During this period changing current induces changing magnetic flux across the primary coil.

This changing magnetic flux is linked across the secondary coil and induces current (e.m.f) in the secondary coil.

However, note that the direction of current in the secondary coil is reversed and hence the deflection in the galvanometer is opposite to the previous case.

Faraday’s Laws of Electromagnetic Induction:

I Law:

Whenever there is a change in the magnetic flux linked with a circuit, an emf and hence a current is induced in the circuit. However, it lasts only so long as the magnetic flux is changing.

II Law:The magnitude of the induced emf is directly proportional to the rate of change of magnetic flux linked with a circuit.

ε α dΦ / dt

ε = (Φ2 – Φ1) / t43Norah Ali Al- moneef

Lenz’s law

Lenz’s Law: The induced emf or current always tends to oppose or cancel the change that caused it.


Currents (I) induced in a wire loop.


Faraday’s Law and Electromagnetic Induction• The instantaneous emf induced in a circuit equals

the time rate of change of magnetic flux through the circuit

• If a circuit contains N tightly wound loops and the flux through each loop changes by ΔΦ during an interval Δt, the average emf induced is given by Faraday’s Law:

tN B


Faraday’s Law and Lenz’ Law

• The minus sign is included because of the polarity of the emf. The induced emf in the coil gives rise to a current whose magnetic field OPPOSES ( Lenz’s law) the change in magnetic flux that produced it


Faraday’s Law• Calculates induced emf due to changing magnetic flux.• Unit of Flux: Weber (Wb = Tm2) B B A

• Changing the flux B = BAcos()– a) Change B– b) Change A– c) Change

Induced Potential

d dB=A cos

dt dtB

d dA=B cos

dt dtB

d cosd

=BAdt dt

B

dt

)d(AB.cos

dt

dBA

dt

dV . B NNNdlE

•The number of loops (N) in the coilNote: magnetic flux changes when either the magnetic field (B), the area (A) or the orientation (cos ) of the loop changes


How can we change the flux?

• Change flux by:– Change area– Change angle– Change field

Bd

dtemf


Changing Area

A loop of wire (N=10) contracts from 0.03 m2 to 0.01 m2 in 0.5 s, where B is 0.5 T and is 0o (Rloop is 1 ).

dA-NB cos

dt

1) What is the induced in the loop?2) What is the induced current in the loop? 51Norah Ali Al- moneef

Changing Orientation

A loop of wire (N=10) rotates from 0o to 90o in 1.5 s, B is 0.5 T and A is 0.02 m2 (Rloop is 2 ).

1) What is the induced in the loop?2) What is the induced current in the loop?

( )

( )

d cos-NAB

dt

d cos ωt-NAB

r

dt

o


Faraday’s Law of Induction

Faraday’s Law: The instantaneous voltage in a circuit (w/ N loops) equals the rate of change of magnetic flux through the circuit:

if

if

ttN

tNV

The minus sign indicates the direction of the induced voltage. To calculate the magnitude:

if

if

ttN

tNV


Lenz’s Law• When the magnetic flux changes within a loop of wire, the induced

current resists the changing flux• The direction of the induced current always produces a magnetic

field that resists the change in magnetic flux (blue arrows)

• Review the previous examples and determine the direction of the current

B

Magnetic flux, B

B

Increasing B

i

B

Increasing B

i


There are three possibilities to produce an emf

1) Time-varying magnetic field=-N[(A cos)(B/t)+

2) Time-varying loop area +(B cos )(A/t)+

3) Turning of the loop (generator) +BA([cos]/t)]


A UHF television loop antenna has a diameter of 11 cm. The magnetic field of a TV signal is normal to the plane of the loop and, at one instant of time, its magnitude is changing at the rate 0.16 T/s. The magnetic field is uniform. What emf is induced in the antenna?

clok wise is

page theinto is

mV -1.52 4

)11.0(14.3 16.0

4

)11.0(14.3

2

).(

2

22

2

22

I

B

md

rA

dt

ABd

dt

d

ind

ind

ind

ind

Example


The magnetic flux through the loop shown in Fig increases according to the relation B = 6.0t 2 + 7.0t, where B is in

milliwebers and t is in seconds. (a) What is the magnitude of the emf induced in the loop when t = 2.0 s? (b) What is the direction of the current through R?

left the toisresistor hrough tI

clock wise I

page theinto B

growing

page ofout B direction

mV 10-3

s 2 at t

Wb/s10)712( mWb/s 712)76(

ind

ind

3

32

dt

d

ttdt

ttd

dt

ddt

dind

example


V101.1 -2

5.0 10-4.5

sec 2 T 5.0)

2-2-

2

tBadt

dBr

dt

dBA

dt

d

V101.1 2

5.0 104.5

sec 2 T 5.0)

2-2-

tBc

The magnetic field through a single loop of wire, 12 cm in radius and of 8.5 resistance, changes with time as shown in the figure. Calculate the emf in the loop as a function of time. Consider the time intervals (a) t = 0 to t = 2.0 s, (b) t = 2.0 s to t = 4.0 s, (c) t = 4.0 s to t = 6.0 s. The (uniform) magnetic field is perpendicular to the plane of the loop.

example

0

sec 2 T 0)

tBb


/ 4.1

4

101.1

V 101.1 101.1 10 R

101.1

4)105.2(

10.01068.1 2)

2

2

23

3

3 2

8

stesladdt

dBdt

dBA

dt

d

I

A

r

A

lRa

Aloop

indind

Example A uniform magnetic field is normal to the plane of a circular loop 10 cm in diameter and made of copper wire (of diameter 2.5 mm). (a) Calculate the resistance of the wire. (see copper resistivity in table) (b) At what rate must the magnetic field change with time if an induced current of 10 A is to appear in the loop?


: In the figure a 120-turn coil of radius 1.8 cm and resistance 5.3 is placed outside a solenoid like that in the figure . The solenoid current drops from 1.5 A to zero in time interval Δt = 25 ms, what current appears in the coil while the solenoid current is being changed?

example


0 2 4 6 8t (s)

0.5

1.0

B (T)

example: What is the current produced in a loop of radius 16 cm and resistance 8.5 at different times shown below? Assume the loop has an orientation in which it receives maximum magnetic flux.


62

Example• Find the emf in a

coil of 200 turns when there is a change of flux of 30 mWb linking it in 40 ms.

• Solution

• Δϕ = 30 x 10-3 Wb• Δt = 40 x 10-3 s

dt

dNE

3

3

1040

1030

x

xNE

Induced emf, EInduced emf, E

VE 150


example: A 50 turn rectangular coil of dimensions 5.0 cm x 10.0 cm is allowed to fall from a position where B = 0 to a new position where B = 0.500 T and is directed perpendicular to the plane of the coil. Calculate the magnitude of the average emf if this occurs in 0.250 seconds.


• : The magnetic field is increasing at a rate of 4.0 mT/s. What is the direction of the current in the wire loop?

example


• The magnetic field is increasing at a rate of 4.0 mT/s. What is the direction of the current in the wire loop?

example


• The magnetic field is decreasing at a rate of 4.0 mT/s. The radius of the loop is 3.0 m, and the resistance is 4 . What is the magnitude and direction of the current?

dt

)d(AB.cos

dt

dBA

dt

dV . B NNNdlE

example


example

• B decreases by 3.00 mT/s in a circular region of radius r =5.00 cm. What is the EMF around the loop?

x x

x x x x

x x x x

x x

externalB

V 003.005.0 2

2

t

Br

Example• A flat circular coil of 600 turns and with a mean radius of 1.5 cm is connected

between the terminals of an ammeter. The total resistance of the coil and the ammeter is 0.5 Ω. The plane of the coil is placed at right angles to a uniform magnetic field of magnetic flux density 0.34 T. The coil is removed from the magnetic field in a time of 60 ms. Calculate the magnitude of :

a) the average induced emf across the ends of the coil

magnetic flux linkage = NΦ = BAN Initial magnetic flux linkage = 600 x 0.34 x (π x 0.0152) = 0.144 WbAfter the coil is removed from the magnetic field, B = 0 therefore final magnetic flux linkage = 0.ε = - (d N Φ / dt) = (0 – 0.144) / 60 x 10-3 = 2.4 V

b) the average induced current in the coil

I = ε / R = 2.4 / 0.5 = 4.8 A


Example• A flat coil of 680 turns and mean cross-sectional area 4.5 x 10-4 m2 is placed in

a region of uniform magnetic field of flux density 580 μT. Initially, the plane of the coil is at right angles to the magnetic field. Calculate the mean induced ε across the ends of the coil when:

a) the coil is withdrawn from the region of field in a time of 60 ms.Initial magnetic flux linkage = NΦ = BAN = 680 x 580 μ x (4.5 x 10-4) = 1.77 x 10-4 WbFinal magnetic flux linkage = 0.ε = - (d N Φ / dt ) = (0 – 1.77 x 10-4) / 60 x 10-3 = 3.0 mV

b) the coil is rotated through 90o in a time of 60 ms.Final magnetic flux linkage = 0, therefore ε = 3.0 mV

c) the direction of the field is reversed in a time of 30 ms.Initial flux = 1.77 x 10-4 WbFinal flux = -1.77 x 10-4 Wb ε = - (d N Φ / dt) = (-1.77 x 10-4 -1.77 x 10-4 ) / 30 x 10-3 = 12 mV


Norah Ali Al- moneef 70

example

• A 0.25 T magnetic field is perpendicular to a circular loop of wire with 50 turns and a radius 15 cm.

• The magnetic field is reduced to zero in 0.12 s.• What is the magnitude of the induced EMF?• Flux = BA = (B)(r2) = (0.25 T)(0.15m)2 = 0.0176 T·m2

• Rate of change of flux = t• fiBA = 0.0176 T·m2

• EMF = N t = 0.0176 T·m2 )/(0.12 s) • EMF = 7.35 T·m2 /s• T·m2 /s = [N / (A·m)] (m2/s) =(N·m)/(A·s) = J/C =V• EMF = 7.35 V

31.2 Motional emf• A straight conductor of length ℓ moves

perpendicularly with constant velocity through a uniform field

• The electrons in the conductor experience a magnetic force– F = q v B

• The electrons tend to move to the lower end of the conductor

ℓℓ

• As the negative charges accumulate at the base, a net positive charge exists at the upper end of the conductor

• As a result of this charge separation, an electric field is produced in the conductor

• Charges build up at the ends of the conductor until the downward magnetic force is balanced by the upward electric force

• There is a potential difference between the upper and lower ends of the conductor


Charges stop moving when:

E BF F

qE qvB

VvB

V vB


• Electron holes continue to move up, but only until FE down equals FB up, at which charge separation ceases.

• As long as the wire keeps moving, there will be a charge separation.

• The magnetic force is doing work to maintain that charge separation.

For a wire of length L, moving in a direction perpendicular to an external B field:Charges stop moving when:

E BF F qE qvB

VvB

V vB


Motional emf, cont.V =Eℓ F=qvB

V=Bℓv, voltage across the conductor – If the motion is reversed, the polarity of the

potential difference is also reversed

FF==qEqE==qq ( (VV//ℓℓ ) ) ==qvBqvB


• A conducting bar sliding with v along two conducting rails under the action of an applied force Fapp. The magnetic force Fm opposes the motion, and a counterclockwise current is induced.

Magnitude of the Motional emf

vBt

xB

t

xBAB

B

B


• Once there is a current flowing through the wire, charges move in 2 directions.– They all move right with

the moving wire.– Current moves up.

• Yet another magnetic force generated, this time to the left.


Motional emf• The emf due to the charge separation exists

whether or not the loop is closed (battery analogy).• If there is a closed loop:

I = vLB/R• The induced current is due to magnetic forces on

moving charges.


• As the bar moves to the right, the magnetic flux through the circuit increases with time because the area of the loop increases

• The induced current must be in a direction such that it opposes the change in the external magnetic flux


Magnetic force (F = IL x B) due to the induced current is toward the left, opposite to velocity v.


• This magnetic force opposes the original velocity of the moving wire.

• The moving wire will slow down and stop.

• Need a constant Fpull to the right to make the contraption work.

• To keep the wire at constant speed, and continue the emf and current Fpull = Fmag

F = ILBFpull = Fmag = ILB

I = vLB/RF = vL2B2/R


Fpull = Fmag

Wpull = Wmag

The rate at which work is done on the circuit equals the power dissipated by the circuit:

P = I2R = v2L2B2/R


Linear Generator with Faraday’s Law Area x

B B dA B A B x

Bd dB x

dt dt

dxB B v

dt

By Lenz’s Law, what is the direction of current?B v

IR


Motional emf in a Circuit, cont.• The changing magnetic

flux through the loop and the corresponding induced emf in the bar result from the change in area of the loop

• The induced, motional emf, acts like a battery in the circuit

R

vBIvB

and


Lenz’ Law, Bar Example, cont

• The flux due to the external field is increasing into the page

• The flux due to the induced current must be out of the page

• Therefore the current must be counterclockwise when the bar moves to the right


Lenz’ Law, Bar Example, final

• The bar is moving toward the left

• The magnetic flux through the loop is decreasing with time

• The induced current must be clockwise to to produce its own flux into the page


Force & Magnetic InductionWhat about the force applied by the hand to keep the rail moving?• The moving rail induces an electric current and also produces power

to drive the current:

P = .i = (5 V)(2.5 A) = 12.5 W• The power (rate of work performed) comes from the effort of the

hand to push the rail– Since v is constant, the magnetic field exerts a resistive force on the rail:

The force of the hand can be determined from the power:

Net hand B hand BF = F + F = 0 or F = F

hand hand

PP = F v F =

v

hand Bms

12.5 WF = = 12.5 N =F

10

BF

handF


Power moving the bar

bar

B vF I B B

R

2 2 2B v B vP B v

R R

2 2 2 2 2 22

2

B v B vP I R R

R R

Same result!


Example:Rod and rail have negligible resistance but the bulb has a resistance of 96 B=0.80 T, v=5.0 m/s and ℓℓ =1.6 m.

Calculate (a) emf in the rod, (b) induced current (c) power delivered to the bulb and (d) the energy used by the bulb in 60 s.

(a) =vBℓ =(5.0 m/s)(0.80 T)(1.6 m)=6.4 V(b) I=/RI=(6.4V)/(96 )=0.067 A(c) P=IP=I=(6.4 V)(0.067 A)=0.43 W(d) E=PtE=(0.43 W)(60 s)=26 J (=26 Ws)


Operating a light bulb with motional EMFConsider a rectangular loop placed within a

magnetic field, with a moveable rail (Rloop= 2 ).

B = 0.5 Tv = 10 m/sL = 1.0 m

Questions:1) What is the induced in the loop?2) What is the induced current in the loop?3) What is the direction of the current?


• A conducting rectangular loop moves with constant velocity v in the +x direction through a region of constant magnetic field B in the -z direction as shown.– What is the direction of the induced current in

the loop?

(a) ccw (b) cw (c) no induced current• A conducting rectangular loop moves with constant velocity v in the -y direction and a constant current I flows in the +x direction as shown.

• What is the direction of the induced current in the loop?

1A

X X X X X X X X X X X XX X X X X X X X X X X XX X X X X X X X X X X XX X X X X X X X X X X X

v

x

y

(a) ccw (b) cw (c) no induced current

1B v

I

x

y



the loop?

• There is a non-zero flux B passing through the loop since B is perpendicular to the area of the loop.• Since the velocity of the loop and the magnetic field are CONSTANT, however, this flux DOES NOT CHANGE IN TIME.• Therefore, there is NO emf induced in the loop; NO current will flow!!

(c) no induced current(a) ccw (b) cw

1A


v

x

y



the loop?

y

• The flux through this loop DOES change in time since the loop is moving from a region of higher magnetic field to a region of lower field.• Therefore, by Lenz’ Law, an emf will be induced which will oppose the change of flux.• The current i is induced in the clockwise direction to restore the flux.

(a) ccw (b) cw (c) no induced current• A conducting rectangular loop moves with constant velocity v in the -y direction and a constant current I flows in the +x direction as shown.

• What is the direction of the induced current in the loop?


v

x

(a) ccw (b) cw (c) no induced current

v

I

x

y i


94

Ans. 4.05 V, 2.03 A.

Example: A coil is wrapped with 200 turns of wire on the perimeter of a square frame of side 18 cm. The total resistance of the coil is 2Ω . B is the plane of the coil and changes linearly from 0 to 0.5 T in 0.80 seconds. Find the emf in the coil while the field is changing. What is the induced current?


Another example: a bar magnet is moved to the left/ right toward a stationary loop of wire.


96

Example Induced EMF, Induced Current.

• A solenoid (similar to one used for a class demonstration) has a diameter of 10 cm, a length of 10 cm, and contains 3500 windings with a total resistance of 60 Ohm.

• The solenoid is connected in a simple loop, modeled above.• Initially, the solenoid is embedded in a magnetic field of

0.100 T, parallel to the axis of the solenoid, as shown.• This external field is reduced to zero in 0.10 sec.• During this 0.1 sec, what is the EMF in the coil, what is the

current in the circuit, and what is the direction and magnitude of the magnetic field in the solenoid generated by this current?


97

Flux & induced EMF

• First, what is the initial flux?– Let us chose current flowing around the circuit in the clockwise

direction to be positive. Such a current would generate a magnetic field pointing up in the solenoid (and pointing down outside the solenoid). Thus the initial flux is positive.

– Flux in one winding of solenoid • = (Area)(Magnetic Field) = r2 B = 3.14 (0.05m)2 (0.100 T).• = 7.85e-4 T m2

• fie4 Tm2 • t = 0.100 sec• EMF = N (/t) = (3500)(7.85e-3 T m2/s) = 27.5 V


98

Induced Current

• EMF in loop = 27.5 V– Changing flux acts just like a battery

• EMF – IR = 0• I = EMF/R = (27.5 V) / 60 =0.458 A

I


example: In the figure, a long rectangular conducting loop, of width L, resistance R, and

mass m, is hung in a horizontal, uniform magnetic field B that is directed into the page and that exists only above line aa. The loop is then dropped; during its fall, it accelerates until it reaches a certain terminal speed vt. Ignoring air drag, find that terminal speed.


A metal rod is forced to move with constant velocity along two parallel metal rails, connected with a strip of metal at one end, as shown in the figure. A magnetic field B = 0.350 T points out of the page. (a) If the rails are separated by 25.0 cm and the speed of the rod is 55.0 cm/s, what emf is generated? (b) If the rod has a resistance of 18.0 and the rails and connector have negligible resistance, what is the current in the rod? (c) At what rate is energy being transferred to thermal energy?

mW 10 1.3R Ior p ) c

clock wise 0027.0

18 R is loopfor resistance total) b

V0lt 05.025.055.035.0

d

Lx A A

4-2

2

ind

R

AR

I

BLVdt

dxBL

dt

B

example:


How much current flows through the resistor? How much power is dissipated by the resistor?

B = 0.15 T

v = 2 m/s

50 cm3

example

Counter clock wise ( direction)

Example

An airplane travels at 1000 km/hr in a region where the earth’s magnetic field is 5 X 10-5T (vertical). What is the potential difference between the wing tips if they are 70 m apart?

1000 km/hr = 280 m/s ε = Blv ε = (5 X 10-5T )(70 m)(280 m/s) = 1.0 V

ExampleBlood contains charged ions. A blood vessel is 2.0 mm in

diameter, the magnetic field is 0.080 T, and the blood meter registers a voltage of 0.10 mV. What is the flow velocity of the blood?

E = Blvv = E /Blv = (1.0 X 10-4 V) (0.080 T)(0.0020m)

v = 0.63 m/s

1norah ali al- moneef example : sketch the path and calculate the acceleration. e=2000 n/c, and v =...

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