A particle of charge $q$ moves with a velocity $\vec{V}=$ ai in a magnetic field $\overrightarrow{\mathrm{B}}=\mathrm{b} \hat{\mathrm{j}}+\mathrm{ck}$, where ' a ', ' b ' and ' c ' are constants. The magnitude of force experienced by particle is

The magnetic force on a charged particle is given by the Lorentz force law: $\vec{F} = q (\vec{V} \times \vec{B})$.

Given $\vec{V} = a\hat{i}$ and $\vec{B} = b\hat{j} + c\hat{k}$, the cross product is computed as:

$\vec{V} \times \vec{B} = (a\hat{i}) \times (b\hat{j} + c\hat{k}) = ab(\hat{i} \times \hat{j}) + ac(\hat{i} \times \hat{k})$.

Using the identities $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{i} \times \hat{k} = -\hat{j}$, the result simplifies to:

$\vec{V} \times \vec{B} = ab\hat{k} - ac\hat{j}$.

Substituting back into the force equation yields:

$\vec{F} = q(ab\hat{k} - ac\hat{j}) = -qac\hat{j} + qab\hat{k}$.

The magnitude is found using the Euclidean norm:

$|\vec{F}| = \sqrt{(-qac)^2 + (qab)^2} = \sqrt{q^2 a^2 c^2 + q^2 a^2 b^2} = |qa| \sqrt{b^2 + c^2}$.

For physical quantities, we take absolute values, giving $|\vec{F}| = qa \sqrt{b^2 + c^2}$, which corresponds to option A.

Final Answer: $\boxed{\text{A}}$