Motion of a body under the influence of increasing magnetic field and a spring

I came across a problem as follows:

A particle of mass $$m$$ and charge $$q$$ moves on a circular path on the surface of a frictionless table with speed $$v$$ where a magnetic field $$B$$ exists uniformly over the whole region. It is attached by a string that passes through a hole in the table to a spring, as shown. The spring is stretched by $$x$$. If now the magnetic field is increased slowly to $$2B$$, what will be the effect on the extension of the spring and the speed of the particle?

My Attempt

Instinctively, it feels that as the magnetic field would increase, the outward radial force would increase, hence the spring's extension. And as the extension increases, the potential energy would increase. Thus the kinetic energy and, therefore, velocity should decrease.

And these are the answers to the question. The extension would increase, and the velocity would decrease.

Doubts

But I have doubts regarding the reasoning.

1. I am arguing that the outward magnetic radial force is increasing, but the centrifugal force is decreasing due to the decreasing velocity and the increasing radius, so the tension and extension of the spring might increase or decrease.

2. For the second part, I feel that tension might also do some work as the object is probably spiralling outwards. So the conservation of mechanical energy might not be possible.

I know that I am contradicting my statements, but I am perplexed about it. I tried writing the mathematical expression and differentiating it but could not get any result as there were many variables involved.

Any help would be greatly appreciated.

• Equations:$\bullet \frac{1}{2} kx^2+\frac{1}{2} mv_0^2=\frac{1}{2} kX^2+\frac{1}{2} mv^2$ $\bullet \frac{mv_0^2}{R}=kX-2qvB$ $\bullet R=X+r-x$ $\bullet \frac{mv_0^2}{r}=kx-qv_0B$ unknowns are X,v and R try solving these Commented Sep 9, 2021 at 7:00