Explanation:
Given that,
Length of the spring, l = 50 cm
Mass, m = 330 g = 0.33 kg
(A) The mass is released and falls, stretching the spring by 28 cm before coming to rest at its lowest point. On applying second law of Newton at 14 cm below the lowest point we get :
(B) The amplitude of the oscillation is half of the total distance covered. So, amplitude is 14 cm.
(C) The frequency of the oscillation is given by :
The forces on one of the wires will be 0.000264 N.It is the field sensed around a moving electric charge that gives rise to the magnetic force.
<h3>What is a magnetic field?</h3>It is the type of field where the magnetic force is obtained. With the help of a magnetic field. The magnetic force is obtained it is the field felt around a moving electric charge.
The given data in the problem is;
I is the current = 6.50 A
B is the magnetic field=f 2.71 X 10⁻⁵ T
The force on the wire is;
Hence, the force on one of the wires will be 0.000264 N.
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