Editorial - Fisika - Induksi Elektromagnetik

General Formulae

Magnetic Flux

B=φ/A

φ=B⋅A

B adalah tegak lurus, use cos(θ) for the angle relative to normal

φ adalah flux (SCALAR)

Electric Force via Induction

E=-n⋅Δ(φ)/Δ(t)=-n⋅Δ(B⋅A)/Δ(t)

E=-n⋅d(φ)/d(t)

E=I⋅R

E=B⋅l⋅v

n jumlah lilitan

Δ(φ) adalah perubahan flux

Δ(t) adalah lamanya terjadi Δ(φ)

Inductance

E=-L⋅Δ(I)/Δ(t)=-L⋅d(I)/d(t)

L=Δ(φ)/Δ(I)

L is the coefficient of inductance

The - is there because it opposes the direction of current.

Lenz

This is for the coil problems with the galvanometer.

(Hukum Lenz): Arah arus induksi selalu menentang perubahan fluks magnetik penyebabnya.

If I move the magnet closer, then that will create a pole with the same polarity in order to push it away, while if I move it away, it will create a pole of the opposite polarity to pull it closer. For each force there is an opposing one.

Like, if I push U-S into a coil, it will create S-U to oppose that pushing force.

Then, use the right hand grip rule:

Have your thumb pointing towards the north pole of the coil.

Then, pay attention to the direction of your fingers that are "behind" and in front of the coil, which shows the direction of current.

Power

W=(L*I2)/2

Generator

(E_max)=n*B*A*ω

Transformer

(V_1)/(V_2)=(n_1)/(n_2)

(V_1)*(I_1)=(V_2)*(I_2)

Above uses ideal transformers.

Prinsip induksi silang.

Left is primary, right is secondary.

Left is input, right is output.

Power stays constant.

Inti besi dibuat berlapis agar tidak panas sehingga meminimalisir kehilangan energi.

Panas terjadi karena pemanasan joule atau panas dari kumparan.

In a non-ideal system (P_2)<(P_1) therefore η=(P_2)/(P_1)⋅100

Step up (V_2)>(V_1) (I_2)<(I_1) (n_2)>(n_1)

Step down (V_2)<(V_1) (I_2)>(I_1) (n_2)<(n_1)

Problems

Problem 1

φ=20⋅10(-3)⋅50⋅10(-4)⋅3/5=3/50000=6/100000=6⋅10(-5)

Problem 2

1 dan 3

Problem 3

d(φ)/d(t)=6*t+2

E=10⋅(6+2)=80

Problem 4

φ=50⋅10(-3)⋅20⋅10(-4)=1/10000=1⋅10(-4)

E=500⋅2⋅(1⋅10(-4))/1/4=2/5=0.4

Multiply by 2 because the direction reverses, therefore it has to go to 0 and then to the opposite "negative" value.

Problem 5

E=B⋅l⋅v=2/10⋅60⋅10(-2)⋅2=6/25

I=6/25⋅1/4=3/50=0.06

Dari P ke Q, (A)

(maksudnya keluar dari P masuk ke Q, gitu ya)

Problem 6

F=2/10⋅6/100⋅60⋅10(-2)=0.0072

Ke kiri, (E)

Problem 7

E=6⋅10(-5)⋅√(,3)/2⋅25⋅140=21/200√(,3)=105√(,3)

(E)

Problem 8

E=n⋅Δ(B⋅A⋅cos(θ))/Δ(t)

The area is what's changing here.

The rotational period p=0.2

E=n⋅B⋅Δ(A)/Δ(t)

E=1⋅0.3⋅3.14⋅((80⋅10(-2))2)/1/5≈3

(C)

Problem 9

E=((10⋅10(-2))2⋅2/10⋅50⋅(4/5-0))/(4⋅10(-3))=20

Correct

Problem 10

(D)

Problem 11

L=200⋅(50⋅10(-3))/2=5

(A)

Problem 12

5⋅10(-3)⋅(2⋅10(-1))/(2⋅10(-3))=1/2

(D)

Problem 13

I=2*sin(100*t)

(I^′)=200*cos(100*t)

E=50⋅10(-3)⋅200⋅1=10

Problem 14

L=200⋅(4⋅10(-2))/2=4

W=(4⋅22)/2=8

Problem 15

ω=50⋅2*π=100*π

(E_max)=20⋅10(-4)⋅500⋅40⋅10(-3)⋅100*π=4*π

(A)

Problem 16

100/v=2/5

v=250

500=250⋅I

I=2

(A)

Problem 17

1 and 4

Problem 18

200/(V_2)=400/20

(V_2)=10

(V_3)=50/20⋅10=25

450=10⋅(I_2)

(I_2)=45

450=25⋅(I_3)

(I_3)=18

(E)

Problem 19

(V_1)/200=500/2000

(V_1)=50

(P_1)=100/80⋅20=25

25=50⋅(I_1)

(I_1)=1/2

(B)

Problem 20

(E)