One long wire carries current 30.0 A to the left along the x axis. A second long wire carries current 50.0 A to the right along the line (y 5 0.280 m, z 5 0). (a) Where in the plane of the two wires is the total magnetic field equal to zero? (b) A particle with a charge of 22.00 mC is moving with a velocity of 150i ^ Mm/s along the line (y 5 0.100 m, z 5 0). Calculate the vector magnetic force acting on the particle. (c) What If? A uniform electric field is applied to allow this particle to pass through this region undeflected. Calculate the required vector electric field.

y = -0.42 m

1.16*10^-4 T(-k)

-1.73*10^4 N/C (j)

Explanation:

(a) Above the pair of wires, The field out of the page of the 50 A current will be stronger than the (—k) field of the 30 A current (k).  

Between the wires, both produce fields into the page.  

below the wires, y = -  | y |

B = u_o*I/2πr (-k) + u_o*I/2πr (k)

0 = u_o/2πr[50/  | y |+0.28 (-k) + 30/| y | (k) ]

50 | y | = 30(| y | + 0.28)

| y | = -y

-50 y = 30*(0.28 - y)

y = -0.42 m

b)  B = u_o*I/2πr (-k) + u_o*I/2πr (k)

B = 4π*10^-7/2π[ 50/0.28 -1 (-k) +30/1(-k) ]

  = 1.16*10^-4 T(-k)

F = qv*B

F = (-2*10^-6)*(150*10^6(i) )(1.16*10^-4(-k))

F = 3.47*10^-2 N(-j)

c) F_e = qE

      E = F_e/q

E = 3.47*10^-2/-2*10-6

  = -1.73*10^4 N/C (j)

  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