Standard Deviation of a Coin Toss

I am trying to calculate the standard deviation of a coin toss where the probability of one coin toss being either a 'head' or a 'tail' is 50%.

I have made the calculations using the following formulas:

μ (Mean)= Nπ (N = number of trials, π = probability of success)

σ2 (Variance) = Nπ(1-π)

σ (Standard Deviation) = (Nπ (1-π))Square Rooted

(I appologise for the way I have expressed the last formula, but I don't know how to show Square Root signs on the forum)

So far I have made calculations for a sample of 12 and a sample of 10,000

Sample of 12:

μ = Nπ= (12)(0.5) = 6

σ2 = Nπ(1-π)= (12)(0.5)(1.0 - 0.5) = 3.0

σ = 1.73

Sample of 10,000:

μ = Nπ= (10,000)(0.5) = 5,000

σ2 = Nπ(1-π)= (10,000)(0.5)(1.0 - 0.5) = 2,500

σ = 50

These calculations don't seem right to me. I would expect the standard deviation to be roughly between 1.05 to 1.10. I would also expect the

standard deviation to become reasonbaly constant once over 10,000 trials have been calculated, but if I calculate the stanard deviation for 1,000,000

trials I get, σ = 500

Where am I going wrong?

Dan