Asolenoid is made from a copper wire that is 15 m long and has a radius of 1.6 mm. the resistivity of copper is 1.68 x 10-8 ωm. to form the solenoid, the copper wire is tightly wrapped around a tube that is 0.35 m long and has a diameter of 2.5 cm. the solenoid is connected to a power source which supplies an emf of 2.5 v. find the magnitude of the magnetic field through the center of this solenoid

The magnitude of the magnetic field through the center of this solenoid is 0.0568 T.

Explanation:

Given that,

Length = 15 m

Radius r₁ = 1.6 mm

Resistivity of copper

Radius

emf = 2.5 V

Length of tube = 0.35 m

We need to calculate the area of cross section

We need to calculate the resistance

Using formula of resistivity

Put the value into the formula

We need to calculate the current

Using formula of current

Where, = emf

Put the value into the formula

We need to calculate the number of turns per unit length

Using formula of number of turns

We need to calculate the magnetic field

Using formula of magnetic field

Hence, The magnitude of the magnetic field through the center of this solenoid is 0.0568 T.

  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