Does an acceleration always perpendicular to velocity imply a circular motion? In a circular motion the acceleration is always perpendicular to the velocity.
But is the converse true?
If yes then is there any rigorous mathematical proof behind it?
This caused a confusion when I learnt about the Lorentz Force and every source I referred quoted that 'As this force is perpendicular to the velocity, it acts a centripetal force and hence it moves in a circle'
So does it always imply that an acceleration perpendicular to the velocity results in a centripetal force?
 A: The condition that acceleration is perpedicular to velocity can be expressed as
$$\vec{a}\cdot\vec{v} = 0$$
Using the definition of acceleration, we can write this as
$$\frac{\textrm{d}\vec{v}}{\textrm{d}t}\cdot\vec{v} = 0$$
$$\frac{1}{2}\frac{\textrm{d}\left(\vec{v}\cdot\vec{v}\right)}{\textrm{d}t} = 0$$
$$\vec{v}\cdot\vec{v} = \textrm{constant}$$
$$||\vec{v}|| = \textrm{constant}$$
which means the speed is constant.
Notice that all of the steps above are reversible. So, the converse is also true: a constant speed also implies that the acceleration is perpendicular to the velocity.
But, nothing in the above implies that the motion will be circular. If the acceleration is of constant magnitude, then you will get circular motion (or helical motion in 3D). However, a non-constant-magnitude acceleration will not result in circular motion. Imagine a car that is coasting at a constant speed. Turning the steering wheel causes the car to turn left or right. The acceleration of the car is perpendicular to the velocity of the car because the car is traveling at a constant speed. The driver can turn the steering wheel at anytime to take any meandering path and it will still be true that the acceleration is always perpendicular to the car's velocity due to the mathematical derivation above.
Your sources for the Lorentz force are assuming the magnetic field is uniform in space and constant in time. There are other kinds of magnetic fields that induce non-ciruclar motion. For example, there is a magnet called a "wiggler" or "undulator" that is used in free-electron lasers. See the picture below:

The black wiggling line is the path of an electron traveling through a magnetic field that alternates up and down. The alternating magnetic field creates undulating motion in the electron beam. Because the acceleration is caused by magnetic fields, the acceleration is always perpendicular to the electrons velocity. However, because the magnetic field isn't uniform, the induced motion is not circular.
A: 
Does an acceleration always perpendicular to velocity imply a circular motion?

No. Acceleration that is perpendicular to velocity changes ONLY the velocity's direction. The speed remains unchanged, only the direction of velocity. The perpendicular (or normal) acceleration changes the trajectory, and that is all. That trajectory, however, can be circular, elliptic, or anything else.
On the other hand, the tangential component of acceleration changes the speed, and not the direction of velocity.
The normal acceleration is given by the following equation: $a_N = \kappa v^2$ where $\kappa$ is the curvature.
If $\kappa$ is constant, then there is circular motion (in that case $\kappa = 1/r$ where $r$ is the radius, and you end up with the familiar $a_c = v^2/r$ equation).
If you're unfamiliar with the notion of curvature, then the simple bottom line is:

*

*Normal acceleration affects the path the object takes, but that path depends on what the normal acceleration is; it is not necessarily circular.

An example would be pitching a ball sideways -- a component of gravity is going to be normal to the velocity of the ball, but you know the trajectory is parabolic.
A: 
In a circular motion the acceleration is always perpendicular to the velocity.

As said by @user256872, It's No! Here some mathematical treatment you can do.
In polar coordinates,
$$\mathbf{v}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}$$
$$\mathbf{a}=(\ddot{r}-r\ddot{\theta})\hat{r}+(r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{\theta}$$
Suppose they are perpendicular then
$$\mathbf{v}\cdot \mathbf{a}=0$$
or
$$\dot{r}(\ddot{r}-r\ddot{\theta})+r\dot{\theta}(r\ddot{\theta}+2\dot{r}\dot{\theta})=0$$
Which doesn't reduce to conditions for circular motion which is $\dot{r}=0$.
