A current of 19.5 mA is maintained in a single circular loop with a circumference of 2.30 m. A magnetic field of 0.710 T is directed parallel to the plane of the loop. What is the magnitude of the torque exerted by the magnetic field on the loop?

0.0058 Nm

Explanation:

Parameters given:

Number of turns, N = 1

Current, I = 19.5 mA = 0.0195 A

Circumference = 2.3 m

Magnetic field, B = 0.710 T

Angle, θ = 90°

Magnetic torque is given as:

τ = N * I * A * B * sinθ

Where A is area

Circumference is given as:

C = 2 * pi * r

=> 2.3 = 2 * pi * r

=> r = 2.3/(2 * 3.142)

r = 0.366 m

Area, A can now be found:

A = pi * r² = 3.142 * 0.366²

A = 0.42 m²

Therefore, torque is:

τ = 1 * 0.0195 * 0.42 * 0.71 * sin90

τ = 0.0058 Nm

  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

lol

explanation: