An infinitely long straight wire is modified such that it includes a circular loop whose plane is perpendicular to the direction of the (otherwise long, straight) wire. The wire is insulated, and carries a current I, and the radius of the circular loop is R. Find the magnitude and direction of the magnetic field B at the center of the loop.

  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

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.