A moving coil galvanometer has N number of turns in a coil of effective it carries a current I. The magnetic field B is radial. The torque acting on the coil is
(A) $N{A^2}{B^2}I$
(B) $NAB{I^2}$
(C) ${N^2}ABI$
(D) $NABI$

Hint: First start with what is torque in general and how it works then try to substitute the general condition with the case of moving coil galvanometer. Use all the information provided in the question in the specific condition of the moving coil galvanometer and get the required answer.

Complete answer:
Let start with what is torque in general.
Torque, $\tau = force \times perpendicular\,dis\tan ce\,between\,the\,forces$
$\tau = F \times b$
Where,
F is force
And b is perpendicular distance between the forces
Torque acting in case of the moving coil galvanometer will be:
$\tau = BIl \times b$
Where, $l \times b$is the area of the moving coil galvanometer(let area be A).

Now from the question, we know;
Number of turns in the moving coil galvanometer is N. Current carrying in the coil is I.

Therefore, substituting all the values we get;
$\tau = NIAB\sin \theta $
where angle formed by direction of magnetic field with normal on the plane of coil be $\theta $.
$\theta = {90^0}$(radial magnetic field given in question)
So, $\sin \theta = \sin {90^0} = 1$

Final torque will be:
$\tau = NIAB$

Hence the correct answer is Option(D).

Note: Here we have find the torque in case of galvanometer and the angle between the direction of magnetic field and the normal drawn on the plane of coil was given ${90^0}$ due to radial magnetic field given in the question. If it is not radial magnetic field then the angle will get changed and hence the required answer.