Editorial - Fisika - Medan Magnet
Induksi Magnet
B=((μ_0)⋅I)/(2*π*a)
Problem 1
B=(4*π⋅10^(-7)⋅5)/(2*π⋅2⋅10^(-2))=1/20000=5⋅10^(-5)
Arahnya ke barat.
(D)
Problem 2
(B_1)=(4*π⋅10^(-7)⋅3)/(2*π⋅2⋅10^(-2))=3/100000
∧ Ke dalam bidang baca
(B_2)=(4*π⋅10^(-7)⋅2)/(2*π⋅4⋅10^(-2))=1/100000
∧ Keluar bidang baca
(∑_^)(B)=3⋅10^(-5)-1⋅10^(-5)= 2⋅10^(-5)
∧ Ke dalam bidang baca
(C)
Problem 3
(C)
Problem 4
B=2(4*π⋅10^(-7)⋅√(,2))/(2*π⋅2⋅10^(-2))=2√(,2)/100000=2√(,2)⋅10^(-5)
(D)
Problem 5
(B_1)=(4*π⋅10^(-7)⋅3)/(2*π⋅3⋅10^(-2))=1/50000
(B_2)=(4*π⋅10^(-7)⋅2)/(2*π⋅4⋅10^(-2))=1/100000
(∑_^)(B)=√(,1/(50000^2)+1/(100000^2))=√(,5)⋅10^(-5)
Problem 6
(B_1)=(B_2)=(4*π⋅10^(-7)⋅2)/(2*π⋅5⋅10^(-2))=1/125000
(∑_^)(B)=√(,(B_1)^2+(B_2)^2)=√(,1/(125000^2)+1/(125000^2))=√(,2)/125000=(8√(,2))/1000000=8√(,2)⋅10^(-6)
Problem 7
(B_1)=(B_2)=(4*π⋅10^(-7)⋅2)/(2*π⋅2⋅10^(-2))=1/50000
(∑_^)(B)=√(,1/(50000^2)+1/(50000^2)+2⋅1/(50000^2)⋅1/2)=√(,3)/50000=(2√(,3))/100000=2√(,3)⋅10^(-5)
Problem 8
3/(5+x)=2/x
3*x=10+2*x
x=10
Kawat Melingkar
(B_o)=N⋅((μ_0)⋅I)/(2*r)
This is the equation for a point at the center of the circle.
The right hand rule is just the first right hand rule, but you swap I and B, so you have your fingers following the curve and direction of the current, and your thumb showing the direction of the magnetic field.
(μ_0)=4*π⋅10^(-7)
I is current/arus
r is radius
N is amount of circles
B=(B_o)⋅(sin^3)(θ)
This is the equation for a point on the axis of a circle.
Problem 1
B=1/2⋅(4*π⋅10^(-7)⋅2)/(2⋅2⋅10^(-2))=π⋅10^(-5)
∧ Masuk bidang baca.
Problem 2
B=1/4⋅(4*π⋅10^(-7)⋅3)/(2⋅π)=1.5⋅10^(-5)
∧ Keluar bidang baca
Problem 3
(B_1)=1⋅(4⋅10^(-7)⋅π)/(2⋅π⋅10^(-2))=1/50000=2⋅10^(-5)
∧ Ke dalam bidang baca
(B_2)=(4*π⋅10^(-7)⋅2)/(2*π⋅π)=1/π/2500000=4/π⋅10^(-5)
∧ Ke luar bidang baca
(B_1)>(B_2)
(∑_^)(B) ke dalam bidang baca
Problem 4
(B_o)=1⋅(4*π⋅10^(-7)⋅5)/(2⋅8⋅10^(-2))=π/80000
sin(θ)=8/10=4/5
B=π/80000⋅(4^3)/(5^3)=π/156250
Problem 5
(B_o)=1⋅(4*π⋅10^(-7)⋅6)/(2⋅3⋅10^(-2))=π/25000
sin(θ)=3/5
B=π/25000⋅(3^3)/(5^3)=27/3125000*π
B-(B_o)=π/25000-27/3125000=-27/3125000+π/25000=3.1*π⋅10^(-5)
Lorentz
Flip all the hands if the charge is negative.
F=B⋅I⋅l⋅sin(θ)
F=B⋅q⋅v
Or use the other right hand rule mentioned in this section but replace I with v
v is velocity
B is magnetic field
R=(m⋅v)/(B⋅q)
This is for the radius of a particle's path.
Problem 1
(B)
Problem 2
F/m=0.5⋅2=1
∧ Masuk bidang baca
Problem 3
F/l=(4*π⋅10^(-7)⋅3)/(2*π⋅3⋅10^(-2))⋅2=1/25000=4/100000=4⋅10^(-5)
∧ Saling tolak karena arah arusnya berbeda
(C)
Problem 4
F/m=(4*π⋅10^(-7))/(2*π)⋅(5/(10⋅10^(-2))⋅2-2/(5⋅10^(-2))⋅5)=-1/50000=-2⋅10^(-5)
F=-2⋅10^(-5)⋅5⋅10^(-2)=-1⋅10^(-6)
(D)
Problem 5
Left hand rule because electrons are negative
(A)
Problem 6
F=B⋅q⋅v
F=(4*π⋅10^(-7)⋅2)/(2*π⋅5⋅10^(-2))⋅2⋅10^(-6)⋅5⋅10^7=1/1250=8⋅10^(-4)
(D)
Problem 7
(B)
Problem 8
UNSOLVABLE, NO SPEED