1. Write dimensions of each given quantity

We need the combination whose dimension is that of momentum.

Momentum: $[p] = MLT^{-1}$

Charge: $[q] = IT$

Current: $[I] = I$

Permeability of vacuum $\mu_0$:

Using $F = \frac{\mu_0 I_1 I_2}{2\pi r}l$ we get $[\mu_0] = \frac{[F][r]}{[I]^2[l]} = \frac{(MLT^{-2})(L)}{I^2(L)} = ML T^{-2} I^{-2}$

So, $[\mu_0] = MLT^{-2}I^{-2}$

2. Check each option

Option A: $\dfrac{qI}{\mu_0}$

$[qI/\mu_0] = \frac{(IT)(I)}{MLT^{-2}I^{-2}}$ $= \frac{I^2T}{MLT^{-2}I^{-2}} = M^{-1}L^{-1}T^3I^4$ This is not momentum.

Option B: $q^2\mu_0 I$

$[q^2\mu_0 I] = (IT)^2(MLT^{-2}I^{-2})(I)$ $= I^2T^2 \cdot MLT^{-2}I^{-2} \cdot I$ $= MLI$ This is not momentum.

Option C: $\dfrac{q\mu_0}{I}$

$[q\mu_0/I] = \frac{(IT)(MLT^{-2}I^{-2})}{I}$ $= MLT^{-1}I^{-2}$ This is not momentum.

Option D: $q\mu_0 I$

$[q\mu_0 I] = (IT)(MLT^{-2}I^{-2})(I)$ $= IT \cdot MLT^{-2}I^{-2} \cdot I$ $= MLT^{-1}$ This is exactly the dimension of momentum.

3. Final answer

The quantity having the dimension of momentum is: $\boxed{q\mu_0 I}$ So, Option D is correct.

4. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.