prgmNEW1

:STOP:NIFTY
:INDUCTION//FLUX

(Ε)=BCOS(Θ)A=BACOS(Θ)
SI UNIT: 1T*M^2= 1 WEBER
FARADAY:
Ε=-N * (ΔΦ)//ΔT)

A CONDUCTING LOOP IS HELD NEAR A PERSON'S HEAD.
WHEN THE CURRENT IN THE LOOP IS CHANGED RAΠDLY,
THE MAGNETIC FIELD IT CREATES CAN CHANGE AT THE RATE
OF RAΠDLY CHANGING MAGNETIC FIELD
3 X ₁₀^(4T/S. {IF THE MAGNETIC FIELD CHANGES OVER AN AREA OF 1.13*₁₀^(2 M^2, WHAT IS THE INDUCED EMF?}
Ε=ΔΦ//ΔT= (A*ΔB)//ΔT =A(ΔB//ΔT)
=(1.13*₁₀^(-2 M^2)(3*₁₀^(4 T//S)
(Ε)=1.13*1-^-2 M^2  *  3 *₁₀^(4 T//S
(Ε)=339V

A BAR MAGNET IS MOVED RAΠDLY TOWARD A 40 TURN
CIRCULAR COIL OF WIRE. AS THE MAGNET MOVES, THE
AVERAGE VALUE OF BCOS(Θ) OVER THE AREA OF THE COIL
INCREASES FROM O. 0125 T TO 0.45 T IN O. 25S IF
THE RADIUS OF THE COIL IS 3.05 CM, & RESISTANCE OF
ITS WIRE IS 3.55 Ω, {FIND THE MAGNITUDE OF (A) THE
INDUCED EMF.}
A) <PIPE>(Ε)<PIPE>=<PIPE>((Φ)INITIAL-ΦFINAL)//ΔT))<PIPE>
ΦINTITIAL= (BCOSΘ)INITIAL*A= 0.0125T*Π*(0.305^2)= 3.65*₁₀^(-5 T*M^2
ΦFINAL=(BCOS(Θ)_*A=(0.45T)*Π*(0.0305M)^2= 1.32 * ₁₀^(-3 T*M^2
Ε<PIPE>=(40)<PIPE>((132E-3 - 3.65E-5)T*M^2))//0.25S))<PIPE> = 0.205V
{B) FIND THE MAG OF THE INDUCED CURRENT.}
I=V//R
LENZ'S LAW
Ε=-N(ΔFLUX//ΔTIME)
AN INDUCED CURRENT ALWAYS FLOWS IN A DIRECTION THAT OPPOSES THE CHANGE THAT CAUSED IT.
Ε=-N(ΔFLUX//ΔTIME)
Φ=BACOSΘ=1.7T*(1.2M)^2

1. THE MAGNETIC FIELD PRODUCED BY AN MRI SOLENOID 2.6 M LONG AND 1.8 M IN DIAMETER IS 2.1 T.
{FIND THE MAGNITUDE OF THE MAGNETIC FLUX THROUGH THE CORE OF THIS SOLENOID.}
Φ=BACOSΘ=(1.7)((Π/4)(1.2)^2))COS0=1.9WB

2. A SINGLE-TURN SQUARE LOOP OF SIDE L IS CENTERED ON THE AXIS OF A LONG SOLENOID. IN ADDITION, THE PLANE OF THE SQUARE LOOP IS PERPENDICULAR TO THE AXIS OF THE SOLENOID. THE SOLENOID HAS 1340 TURNS PER METER AND A DIAMETER OF 6.00 CM , AND CARRIES A CURRENT OF 2.60 A.
{A. FIND THE MAGNETIC FLUX THROUGH THE LOOP WHEN L= 2.90 CM.}
E IS CONFINED WITHIN THE SOLENOID. AREA WILL BE THE SMALLER OF THE CROSS-SECTIONAL AREA OF THE SOLENOID OR AREA OF THE RECTANGLE.
Φ=BA=(3.926MT)(0.0290M)^2
{B. FIND THE MAGNETIC FLUX THROUGH THE LOOP WHEN L= 6.00 CM}
C. Φ=BA=(3.926MT)Π(0.0290M)^2
{FIND THE MAGNETIC FLUX THROUGH THE LOOP WHEN L= 11.0 CM.}
Φ IS THE SAME AS THAT FOUND IN PART B

3.A COIL WITH MANY LOOPS IS ORIENTED PERPENDICULAR TO A CHANGING MAGNETIC FIELD.
THE AREA OF A 120 TURN COIL ORIENTED WITH ITS PLANE PERPENDICULAR TO A 0.33 T MAGNETIC FIELD IS 4.1×₁₀^(-2 M2.
{FIND THE AVERAGE INDUCED EMF IN THIS COIL IF THE MAGNETIC FIELD REVERSES ITS DIRECTION IN 0.25 S}
A. COIL WITH MANY LOOPS PERPENDICULAR TO A CHANGING MAGNETIC FIELD
<PIPE>Ε<PIPE>=N*<PIPE>ΔΦ//ΔT<PIPE>=N*<PIPE>(BA-(-BA)//ΔT)<PIPE>=N<PIPE>(2BA/ΔT)<PIPE>=120(2(0.33T)((0.050M^2))//0.11=36V

4.A SQUARE LOOP OF WIRE IS IMMERSED IN A MAGNETIC FIELD. THE SQUARE IS RAΠDLY CHANGED INTO A CIRCLE WITH THE SAME PERIMETER.
AN EMF IS INDUCED IN A CONDUCTING LOOP OF WIRE 1.48 M LONG AS ITS SHAPE IS CHANGED FROM SQUARE TO CIRCULAR.
{FIND THE AVERAGE MAGNITUDE OF THE INDUCED EMF IF THE CHANGE IN SHAPE OCCURS IN 0.165 S AND THE LOCAL 0.531-T MAGNETIC FIELD IS PERPENDICULAR TO THE PLANE OF THE LOOP.}
A. C=2ΠR R=1.40M//2Π=0.223M
B. ΦCIRCLE=BA=BΠR^2=(0.335T)Π(0.223M)^2=0.0710WB
C. P=4S S=P//4=1.40//3=0.350M
D.ΦSQUARE=BA=BS^2=(0.455T)(0.350M)^2=0.0557WB
E. Ε=N<PIPE>(ΦCIRCLE-ΦSQUARE)//ΔT)<PIPE>=(1)(0.0710-0.0557WB)//(0.125S)=0.122V

5. A CIRCUIT CONTAINS A RESISTOR AND CAPACITOR. A MAGNETIC FIELD POINTS INTO THE PAGE AND BECOMES STRONGER OVER A SHORT TIME INTERVAL.
SHOWS A CIRCUIT WITH AN AREA OF 0.070 M2 CONTAINING A R=1.0Ω RESISTOR AND A C=280ΜF UNCHARGED CAPACITOR. POINTING INTO THE PLANE OF THE CIRCUIT IS A UNIFORM MAGNETIC FIELD OF MAGNITUDE 0.17T. IN 1.0×102S THE MAGNETIC FIELD STRENGTHENS AT A CONSTANT RATE TO BECOME 0.90T POINTING INTO THE PLANE.
WHAT MAXIΜM CHARGE (SIGN AND MAGNITUDE) ACCΜLATES ON THE UPPER PLATE OF THE CAPACITOR IN THE DIAGRAM?
1. {CALCULATE THE EMF}
E=N<PIPE>ΔΦ//ΔT=(1)ΔBA//ΔT=(0.80-0.15T)(0.060M^2)/0.010S=3.9V
2. {CALCULATE THE CHARGE ON THE UPPER PLATE OF THE CAPACITOR}
Q=-CV=C<PIPE>Ε<PIPE>=-(250*₁₀^(-6F)(3.9)=-975ΜC=-.98MC

6. A FRICTIONLESS ROD SLIDES ACROSS TWO CONDUCTING RAILS THAT ARE PERPENDICULAR TO A UNIFORM MAGNETIC FIELD.
A ZERO-RESISTANCE ROD SLIDING TO THE RIGHT ON TWO ZERO-RESISTANCE RAILS SEPARATED BY THE DISTANCE L = 0.362 M . THE RAILS ARE CONNECTED BY A 10.7 Ω RESISTOR, AND THE ENTIRE SYSTEM IS IN A UNIFORM MAGNETIC FIELD WITH A MAGNITUDE OF 0.680 T .
A) {FIND THE FORCE THAT ΜST BE EXERTED ON THE ROD TO MAINTAIN A CONSTANT CURRENT OF 0.154 A IN THE RESISTOR.}
F=IBL -.0656N
B)WHAT IS THE RATE OF ENERGY DISSIPATION IN THE RESISTOR?
P=I^2R=.306W
C)WHAT IS THE MECHANICAL POWER DELIVERED TO THE ROD?
P=FV=0.306W

7. A FRICTIONLESS ROD SLIDES ACROSS TWO CONDUCTING RAILS THAT ARE PERPENDICULAR TO A UNIFORM MAGNETIC FIELD. THE INDUCED EMF DRIVES
A CURRENT THROUGH A LIGHTBULB. THE LIGHTBULB IN THE CIRCUIT SHOWN IN THE FIGURE HAS A RESISTANCE OF 10 Ω AND CONSUMES 4.9 W OF POWER; THE ROD IS 1.34 M LONG AND MOVES TO THE LEFT WITH A CONSTANT SPEED OF 3.4 M/S . THE STRENGTH OF THE MAGNETIC FIELD IS 1.5 T .
A) {FIND THE CURRENT THAT FLOWS IN THE CIRCUIT.}
I=√(P/R)=.64A
B) {WHAT SPEED ΜST THE ROD HAVE IF THE CURRENT IN THE CIRCUIT IS TO BE 1.1 A ?}
V=Ε/BL=IR/BL=4.1M/S
:STOP