Ignoring spin, consider an electron in a hydrogen 2p orbital, what is its orbital magnetic moment?
The first equation you have is for spin. Let's look at it in a operator form

To properly treat the upcoming discussion, quantum electrodynamics is introduced to describe the eigenstates of spin. The eigenstates of spin has a direct correlation with spin. Formally we depict it as the following

$$ μ_B =-g \frac {eℏ}{2m}  \frac { \boldsymbol {S}}{ℏ} $$

where μ is the magnetic moment operator, g is the g factor, ℏ is plank’s constant over 2π, m is the mass of the particle and S is the spin operator.



For orbital magnetic moment (your question) you use: 

$$\mu =\hbar\sqrt{l(l+1)} \boldsymbol {\mu_B} $$

Recall: 

s: l=0 
p: l=1
d: l=2
f: l=3

So plugging in:

$$\mu =\hbar\sqrt{1(1+1)} \boldsymbol {\mu_B} $$
$$\mu =\hbar\sqrt{2} \boldsymbol {\mu_B}$$

(Units are $ \boldsymbol {\mu_B}$)