Complete question:
At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.
[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]
Answer:
The time it will take the particle to pass through point (P) again is 1.639 ns.
Explanation:
F = qvB
Also;

solving this two equations together;

where;
m is the mass of electron = 9.11 x 10⁻³¹ kg
q is the charge of electron = 1.602 x 10⁻¹⁹ C
B is the strength of the magnetic field = 3.47 x 10⁻³ T
substitute these values and solve for t

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.
Answer:
the object has least potential energy at mean position of the SHM
Explanation:
If a block is connected with a spring and there is no resistive force on the system
In this case the total energy of the system is always conserved and it will change from one form to another form
So here we will say that
Kinetic energy + Potential energy = Total Mechanical energy
As we can say that total energy is conserved so here we have least potential energy when the system has maximum kinetic energy
So here we also know that at mean position of the SHM the system has maximum speed and hence maximum kinetic energy.
So the object has least potential energy at mean position of the SHM
C is the answer hope that helps you
I don't really know the answer but maybe north pole and south pole?