The component of the external magnetic field along the central axis of a 18 18 turn circular coil of radius 27.0 27.0 cm decreases from 1.90 1.90 T to 0.500 0.500 T in 2.90 2.90 s. If the resistance of the coil is R = 4.50 R=4.50 Ω, what is the magnitude of the induced current in the coil?

Induced current I = 0.44 A

Explanation:

given data

no of turn N = 18

radius  = 27.0 cm

resistance of the coil = 4.50 Ω

time = 2.90 s

magnetic field  = 1.90 T to 0.500 T

solution

we get Induced EMF that is

E = - dφ ÷ dt    ..............1

E = - N × A × (B2 - B1) ÷ Δt

E = N × A × (B1 - B2) ÷ Δt

area of cross section of the coil A = π × r²

area of cross section = 3.14 × (0.27)²

area of cross section = 0.228 m²

Initial magnetic field B1   = 1.9 T

and  

Final magnetic field B2   = 0.5 T

time      Δt = 3.7 s

put here value and we get

E  =    

E  = 1.98 V

and  

Induced current I is express as

Induced current I = E  ÷ R    ................2

Induced current I =

Induced current I = 0.44 A

  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

the temperature and the pressure affect solubility. temperature affects by changing the properties of the solvent. the increase in pressure increases solubility, the decrease in pressure decreases solubility.