A sinusoidal voltage V(t) = 100sin(500t) is applied across a pure inductance of L = 0.02H. The current through the coil is :

We know that voltage across inductor
V = L(di/dt)
di = Vdt/L
Integrating both sides,
i = ∫Vdt/L + C
Given V(t) = 100sin(500t) and L = 0.02H
i = ∫100sin(500t)dt / 0.02 + C
i = (100/-500*cos(500t))/0.02 + C
i = -10cos(500t) + C
Assuming initial current is zero, i.e., at t = 0, i = 0
0 = -10cos(0) + C
C = 10
Therefore, i = -10cos(500t) + 10
However, this option is not available. Let's check our integration again.
i = ∫Vdt/L = ∫100sin(500t)dt/0.02
= (100/0.02)∫sin(500t)dt
= 5000 * [-cos(500t)/500] + C
= -10cos(500t) + C
If we assume the initial current is zero, then at t=0, i=0
0 = -10cos(0) + C
=> C = 10
Therefore, i(t) = -10cos(500t) + 10
Let's assume the initial condition is that the current is zero at t = 0.
Then, i(t) = -10cos(500t) + 10. This isn't one of the options. Let's re-examine the problem.
The voltage across an inductor is given by V = L(di/dt).
So, di/dt = V/L = (100sin(500t))/0.02 = 5000sin(500t)
i = ∫5000sin(500t)dt = -10cos(500t) + C
Assuming zero initial current (i.e., at t=0, i=0), then C = 10. So, i(t) = 10 - 10cos(500t).
This solution is still not in the options. The closest option is -10cos(500t). It is possible there is an error in the question or the options provided.