A segment of wire of total length 3.0 m carries a 15-A current and is formed into a semicircle. Determine the magnitude of the magnetic field at the center of the circle along which the wire is placed. a. 1.6 mu T b. 4.9 mu T c. 1.0 mu T d. 9.8 mu T e. 15 mu T

c. 1.0 mu T

Explanation:

From, the ampere's  circuit law:

∫B.dl = μ₀ I

B.(2πr) = μ₀ I

where,

B = Magnetic Field = ?

r = radius of the circle = length of wire = 3 m

μ₀ = Permeability of free space = 4π x 10⁻⁷ Tm/A

I = Current = 15 A

Therefore,

B(2π)(3 m) = (4π x 10⁻⁷ Tm/A)(15 A)

B = 10 x 10⁻⁷ T

B = 1 x 10⁻⁶ T = 1 μT

Hence, the correct option is:

c. 1.0 mu T

  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

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