A solenoid is formed by winding 23.0 m of insulated silver (resistivity 1.59 × 10-8 Ω·m) wire around a hollow cylinder. The turns are wound as closely as possible without overlapping, and the insulating coat on the wire is negligibly thin. When the solenoid is connected to an ideal (no internal resistance) 3.00-V battery, the magnitude of the magnetic field inside the solenoid is found to be 6.69 × 10-3 T. Determine the radius of the wire. Hint: Because the solenoid is closely coiled, the number of turns per unit length depends upon the radius of the wire.

Explanation:

Given that,

Resistivity p = 1.59 × 10^-8 Ω·m

Length of coil L = 23m

Voltage across solenoid V = 3V

Magnetic field B = 6.69 × 10^-3 T.

Radius of wire r=?

Magnetic field is give as

B = μo•i•n

Where n is number of turns per length, then, n= N/L

N = L/2r

Therefore, n = 1/2r

So, B = μo•i/2r

Then, r = μo•i/2B

From ohms law

V=iR

i= V/R

So, r = μo•V/2R•B

Our resistance is given as

R = pL/A

Then, we have

r = μo•V/2R•B

r = μo•V•A/2•p•L•B

A is the area and it is given as

A = πr²

r = μo•V•πr²/2•p•L•B

Then, making r subject of formula

r = 2•p•L•B / μo•V•π

r = (2×1.59×10^-8×23×6.69×10^-3) / (4π×10^-7 × 3× π)

r = 4.13 × 10^-4m

The radius of the wire is 4.13×10^-4m

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  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂