A15.0 cm diameter circular loop of wire is placed with the plane of the loop parallel to the uniform magnetic field between the pole pieces of a large magnet. when 5.30 a flows in the coil, the torque on it is 0.135 m⋅n. what is the magnetic field strength ?

0.0144 tesla

Explanation:

Magnetic field strength is the same as Magnetomotive force

B= I/2πr

B=magnetic field strength

I=current

r=radius=0.15/2=0.75m

Area=πr2= 1.77m2

torqueT= NIABsin theta

B= T/NIABsin theta

B=0.135/1 * 5.3 * 1.77 *sin90

B=0.0144 tesla

B=1.44 tesla

Explanation:

Explanation:

Magnetic field strength is the same as Magnetomotive force

B= I/2πr

B=magnetic field strength

I=current

r=radius=0.15/2=0.075m

Area=πr2= 0.0177m2

torqueT= NIABsin theta

B= T/NIABsin theta

B=0.135/1 * 5.3 * 0.0177 *sin90

B=1.44 tesla

  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

b.a device that inputs heat into a cool object