A velocity selector uses a fixed electric field of magnitude E and the magnetic field is varied to select particles of various energies. If a magnetic field of magnitude B is used to select a particle of a certain energy and mass, what magnitude of magnetic field is needed to select a particle of equal mass but twice the energy?

we need B/√2 for having twice the energy i.e 0.17B

Explanation:

This is actually a simple one, so i will guide you through it.

We know that Energy here means Kinetic Energy  (K.E)

and the expression is given thus;

Energy = 1/2 mv2

and velocity v = E/B

let us make x the energy intially, foe electric field E, magnetic field B, mass m

so, x = 1/2 m (E/B)2

we have to find magnetic field for which twice the energy so , let magnetic field be y, so

2x = 1/2m(E/y)2

y = B/√2  = 0.17B

therefore, we need B/√2 for having twice the energy

cheers i hope this helps

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete