A long, thin solenoid has 155 turns per meter and radius 1 m. The current in the solenoid is increasing at a uniform rate of 28
A/s.
What is the magnitude of the induced electric field at a point 9.5 m from the axis of the solenoid?
Express your answer in mV/m to 2 decimal places.
What is the magnitude of the induced electric field at a point 9.5 m from the axis of the solenoid?
Express your answer in mV/m to 2 decimal places.
1 answer:
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Answer:
Explanation:
mass attached m = .14 kg
force constant k = 5N / m
displacement
= amplitude of oscillation
A = .03 m
A ) period of motion =
= 2 x 3.14
T = 1.05 s
B ) maximum speed of block = angular velocity x amplitude
= (2π /T) x A
= (2 x 3.14 x .03) / 1.05
= .1794 m / s
17.94 cm /s
C )
maximum acceleration = angular velocity² x amplitude
= (2π /T)² x A
= (2π /1.05)² x .03
= 1.073 m / s²
D )
position
S = A cos ωt , ω is angular velocity
S = .03 cos(2πt /T)
= .03 cos 5.98 t
v =ω A sin(2πt /T)
= 5.98 x .03 sin5.98t
= .1794 sin5.98t
acceleration = ω²A sin5.98t
= 1.073 sin5.98t