Answer:
[tex]0.2\; {\rm T}[/tex], assuming that the speed of the electron stays the same.
Explanation:
Let [tex]v[/tex] denote the speed of this electron. Let [tex]q[/tex] denote the electric charge on this electron. Let [tex]m[/tex] denote the mass of this electron.
Since the path of this electron is a circle (not a helix,) this path would be in a plane normal to the magnetic field.
Let [tex]B[/tex] denote the strength of this magnetic field. The size of the magnetic force on this electron would be:
[tex]F = q\, v\, B[/tex].
Assuming that there is no other force on this electron. The net force on this electron would be [tex]F = q\, v\, B[/tex]. By Newton's Second Law of motion, the acceleration of this electron would be:
[tex]\begin{aligned}a &= \frac{F}{m} \\ &= \frac{q\, v\, B}{m}\end{aligned}[/tex].
On the other hand, since this electron is in a circular motion with a constant speed:
[tex]\begin{aligned} a = \frac{v^{2}}{r} \end{aligned}[/tex].
Combine the two equations to obtain a relationship between [tex]r[/tex] (radius of the path of the electron) and [tex]B[/tex] (strength of the magnetic field:)
[tex]\begin{aligned}\frac{q\, v\, B}{m} = \frac{v^{2}}{r}\end{aligned}[/tex].
Simplify to obtain:
[tex]\begin{aligned}r &= \frac{m\, v^{2}}{q\, v\, B} \\ &= \frac{m\, v}{q\, B} \\ &= \left(\frac{m\, v}{q}\right)\, \frac{1}{B}\end{aligned}[/tex].
In other words, if the speed [tex]v[/tex] of this electron stays the same, the radius [tex]r[/tex] of the path of this electron would be inversely proportional to the strength [tex]B[/tex] of the magnetic field. Doubling the radius of this path would require halving the strength of the magnetic field (to [tex]0.2\; {\rm T}[/tex].)
