(a) We know that in the magnetic field

magnetic force = Centrifugal force

\(\displaystyle{q}{V}{B}={\frac{{{m}{V}^{{2}}}}{{{R}}}}\)

\(\displaystyle{R}={\frac{{{m}{V}}}{{{q}{B}}}}\)

Where R is radius

m is mass of helium \(\displaystyle={6.64}\cdot{10}^{{-{27}}}\)

V is speed \(\displaystyle={35.6}\cdot{10}^{{3}}\frac{{m}}{{s}}\)

q is charge on particle =\(\displaystyle{2}\cdot{1.6}\cdot{10}^{{-{19}}}\) C

B is magnetic field=1.1 T

\(\displaystyle{R}={\frac{{{6.64}\cdot{10}^{{-{27}}}\cdot{35.6}\cdot{10}^{{3}}}}{{{2}\cdot{1.6}\cdot{10}^{{-{19}}}\cdot{1.1}}}}={6.72}\cdot{10}^{{-{4}}}\)

Now the diameter (D)=2R=\(13.44 \cdot 10^{-4}\) m

(b) Magnetic field has no effect on speed but it bound the particle to move in circular motion.

(c) We know that radial acceleration \(\displaystyle{\left({a}\right)}={\frac{{{V}^{{2}}}}{{{R}}}}\)

\(\displaystyle={\frac{{{\left({35.6}\cdot{10}^{{3}}\right)}^{{2}}}}{{{6.72}\cdot{10}^{{-{4}}}}}}={1.89}\cdot{10}^{{{12}}}\frac{{m}}{{s}^{{2}}}\)

(d) The direction of the acceleration will be toward center.