A single-turn square loop of wire, 2.00 cm on each edge, carries a clockwise current of 0.240 A. The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30.0 turns/cm and carries a clockwise current of 15.0 A. Find: a. the force on each side of the loop
b. the torque acting on the loop.

Explanation:

a )

Magnetic field inside solenoid B = μ₀ NI ,

μ₀ = 4π x 10⁻⁷ ; N is no of turns per meter length in solenoid and I is current B=  4π x 10⁻⁷ x 30 x 10² x 15

= .0565 T .

Force on each side of square loop = B i L

B is external magnetic field , i is current in loop and L is length of side

Force on each side of square loop = .0565 x .24 x 2 x 10⁻²

= 2.7 x 10⁻⁴ N .

b )

Torque on the loop =  F x d

F is force on one side , d is distance between two sides , that is side of the square loop

= 2.7 x 10⁻⁴ x 2 x 10⁻² N.m

= 5.4  x 10⁻⁶ N.m .

d.7.5cm is your answer

  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