Blue light can knock electrons off a plate, but red light can't
Answer:
0.54 A
Explanation:
Parameters given:
Number of turns, N = 15
Area of coil, A = 40 cm² = 0.004 m²
Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T
Time interval, Δt = 2 secs
Resistance of the coil, R = 0.2 ohms
To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:
|V| = |(-N * ΔB * A) /Δt)
|V| = | (-15 * 3.6 * 0.004) / 2 |
|V| = 0.108 V
According to Ohm's law:
|V| = |I| * R
|I| = |V| / R
|I| = 0.108 / 0.2
|I| = 0.54 A
The magnitude of the current in the coil of wire is 0.54 A
As we know that KE and PE is same at a given position
so we will have as a function of position given as

also the PE is given as function of position as

now it is given that
KE = PE
now we will have




so the position is 0.707 times of amplitude when KE and PE will be same
Part b)
KE of SHO at x = A/3
we can use the formula

now to find the fraction of kinetic energy



now since total energy is sum of KE and PE
so fraction of PE at the same position will be


Data Analysis and Conclusion