Which quantity can be calculated using the equation E = mc2?
2 answers:
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Answer:
The magnitude of the induced electric field at a point 2.5 cm from the axis of the solenoid is 8.8 x 10⁻⁵ V/m
Explanation:
given information:
radius, r = 2.0 cm
N = 700 turns/m
decreasing rate, dI/dt = 9.0 A/s
the magnitude of the induced electric field at a point 2.5 cm (r = 2.5 cm = 0.025 m) from the axis of the solenoid?
the magnetic field at the center of solenoid
B = μ₀nI
where
B = magnetic field (T)
μ₀ = permeability (1.26× 10⁻⁶ T.m/A)
n = the number turn per unit length (turn/m)
I = current (A)
dB/dt = μ₀n dI/dt (1)
now we calculate the induced electric field by using
E =
= 2E/r (2)
where
E = the induced electric field (V/m)
we substitute the firs and second equation, thus
dB/dt = μ₀n dI/dt
2E/r = μ₀n dI/dt
E = (1/2) r μ₀n dI/dt
= (1/2) (0.025) (1.26× 10⁻⁶) (700) (8)
= 8.8 x 10⁻⁵ V/m
Answer:
21.8°
Explanation:
Let's call θ the angle between BC and the horizontal.
Draw a free body diagram for each block.
There are 4 forces acting on block D:
Weight force P pulling down,
Normal force N₁ pushing perpendicular to AB,
Friction force N₁μ pushing parallel up AB,
and tension force T pushing parallel up AB.
There are 4 forces acting on block E:
Weight force P pulling down,
Normal force N₂ pushing perpendicular to BC,
Friction force N₂μ pushing parallel to BC,
and tension force T pulling parallel to BC.
Sum of forces on D in the perpendicular direction:
∑F = ma
N₁ − P sin θ = 0
N₁ = P sin θ
Sum of forces on D in the parallel direction:
∑F = ma
T + N₁μ − P cos θ = 0
T = P cos θ − N₁μ
T = P cos θ − P sin θ μ
T = P (cos θ − sin θ μ)
Sum of forces on E in the perpendicular direction:
∑F = ma
N₂ − P cos θ = 0
N₂ = P cos θ
Sum of forces on E in the parallel direction:
∑F = ma
N₂μ + P sin θ − T = 0
T = N₂μ + P sin θ
T = P cos θ μ + P sin θ
T = P (cos θ μ + sin θ)
Set equal:
P (cos θ − sin θ μ) = P (cos θ μ + sin θ)
cos θ − sin θ μ = cos θ μ + sin θ
1 − tan θ μ = μ + tan θ
1 − μ = tan θ μ + tan θ
1 − μ = tan θ (μ + 1)
tan θ = (1 − μ) / (1 + μ)
Plug in values:
tan θ = (1 − 0.4) / (1 + 0.4)
θ = 23.2°
∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.
