A particle of mass 1.1122 × 10−25 kg and charge of 3.2×10−19 C is accelerated from rest in the plane of the page through a potential difference of 287 V between two parallel plates as shown. The particle is injected through a hole in the right-hand plate into a region of space containing a uniform magnetic field of magnitude 0.0711 T. The particle curves in a semicircular path and strikes a detector. Which way does the magnetic field point?

The magnets would be | n s | and | s n|

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

Ithink the answer is d