I can't find the attached circuit.
Maximum power transfer occurs when the load impedance
is equal to the internal impedance of the power source.
Answer:
A) μ = A.m²
B) z = 0.46m
Explanation:
A) Magnetic dipole moment of a coil is given by; μ = NIA
Where;
N is number of turns of coil
I is current in wire
A is area
We are given
N = 300 turns; I = 4A ; d =5cm = 0.05m
Area = πd²/4 = π(0.05)²/4 = 0.001963
So,
μ = 300 x 4 x 0.001963 = 2.36 A.m².
B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;
B = (μ_o•μ)/(2π•z³)
Let's make z the subject ;
z = [(μ_o•μ)/(2π•B)] ^(⅓)
Where u_o is vacuum permiability with a value of 4π x 10^(-7) H
Also, B = 5 mT = 5 x 10^(-6) T
Thus,
z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)
Solving this gives; z = 0.46m =
Answer:
6) 2.6 m/s, 31°
7) 9.2 m/s
8) 1.2 s
Explanation:
I'll do #6, #7, and #8 as examples. You can solve #9 using the equation from #7, and #10 using the equation from #8.
6) Take north to be +y and east to be +x.
Given:
vₓ = 2.2 m/s
vᵧ = 1.3 m/s
Find: v
v² = vₓ² + vᵧ²
v² = (2.2 m/s)² + (1.3 m/s)²
v ≈ 2.6 m/s
θ = atan(vᵧ / vₓ)
θ = atan(1.3 / 2.2)
θ ≈ 31°
7) Given:
Δy = -4.3 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-4.3 m)
v ≈ 9.2 m/s
8) Given:
Δy = -6.7 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
-6.7 m = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 1.2 s