In designing a velocity selector that uses uniform perpendicular electric and magnetic fields, you want to select positive ions of charge that are traveling perpendicular to the fields at 8.75 km s. the magnetic field available to you has a magnitude of 0.550 t. (a) what magnitude of electric field do you need? (b) show how the two fields should be oriented relative to each other and to the velocity of the ions.

Part a)

E = 4812.5 N/C

Part b)

Two field must be perpendicular to each other as well as perpendicular to the velocity of charge so that net force on the moving charge will be zero

Explanation:

For uniform electric and magnetic field we will have a charge particle moving through it such that net force on it is zero

so here we have

magnetic force on moving charge will balanced by electrostatic force on moving charge

now we have

B = 0.550 T

now we have

Part b)

Two field must be perpendicular to each other as well as perpendicular to the velocity of charge so that net force on the moving charge will be zero

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz