Answer:
Incomplete questions
Let assume we are asked to find
Calculate the induced emf in the coil at any time, let say t=2
And induced current
Explanation:
Flux is given as
Φ=NAB
Where
N is number of turn, N=1
A=area=πr²
Since r=2cm=0.02
A=π(0.02)²=0.001257m²
B=magnetic field
B(t)=Bo•e−t/τ,
Where Bo=3T
τ=0.5s
B(t)=3e(−t/0.5)
B(t)=3exp(-2t)
Therefore
Φ=NAB
Φ=0.001257×3•exp(-2t)
Φ=0.00377exp(-2t)
Now,
Induce emf is given as
E= - dΦ/dt
E= - 0.00377×-2 exp(-2t)
E=0.00754exp(-2t)
At t=2
E=0.00754exp(-4)
E=0.000138V
E=0.138mV
b. Induce current
From ohms laws
V=iR
Given that R=0.6Ω
i=V/R
i=0.000138/0.6
i=0.00023A
i=0.23mA
To reach a vertical height of 13.8 ft against gravity, which has an acceleration of 32 ft/s^2, the required vertical speed can be calculated from the equation:
vi^2 - vf^2 = 2*g*h
Given that it has vf = 0 (it is not moving vertically at its maximum height), g = 32, and h = 13.8, we can solve for vi:
vi^2 = 29.72 ft/s
This is only its vertical speed, so this is equivalent to its original speed multiplied by the sine of the angle:
29.72 ft/s = (v_original)*(sin 42.2<span>°</span>)
v_original = 44.24 ft/s
Converting to m/s, this can be divided by 3.28 to get 13.49 m/s.
The free-body diagram is missing, but I assume the only forces acting on the box are the force F pushing the box, the weight of the object and the normal reaction of the surface.
Since the weight and the normal reaction acts in the vertical (y) direction, the only force acting on the box in the horizontal (x) direction is the horizontal component of the force F, which is given by
And so this is the net force in the x-direction.
Answer:
ΔΦ = -3.39*10^-6
Explanation:
Given:-
- The given magnetic field strength B = 0.50 gauss
- The angle between earth magnetic field and garage floor ∅ = 20 °
- The loop is rotated by 90 degree.
- The radius of the coil r = 19 cm
Find:
calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.
Solution:
- The change on flux ΔΦ occurs due to change in angle θ of earth's magnetic field B and the normal to circular coil.
- The strength of magnetic field B and the are of the loop A remains constant. So we have:
Φ = B*A*cos(θ)
ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )
- The initial angle θ_1 between the normal to the coil and B was:
θ_1 = 90° - ∅
θ_1 = 90° - 20° = 70°
The angle θ_2 after rotation between the normal to the coil and B was:
θ_2 = ∅
θ_2 = 20°
- Hence, the change in flux can be calculated:
ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )
ΔΦ = -3.39*10^-6
