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1 answer:
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Answer:
Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T
Explanation:
Given;
number of turns of solenoid, N = 269 turn
length of the solenoid, L = 102 cm = 1.02 m
radius of the solenoid, r = 2.3 cm = 0.023 m
current in the solenoid, I = 3.9 A
Magnitude of the magnetic field inside the solenoid near its centre is calculated as;
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T
Answer:
Frequency of oscillation, f = 4 Hz
time period, T = 0.25 s
Angular frequency,
Given:
Time taken to make one oscillation, T = 0.25 s
Solution:
Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:
f =
f =
Frequency of oscillation, f = 4 Hz
The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.
Therefore, time period, T = 0.25 s
Angular frequency of oscillation is given by: