Two coils are wound around the same cylindrical form. when the current in the first coil is decreasing at a rate of -0.250 a/s, the induced emf in the second coil has a magnitude of 1.65 x 10^-3 v. a) what is the mutual inductance of the pair of coils? b) if the second coil has 30 turns, what is the flux through each turn when the current in the first coil equals 1.25 a? c) if the current in the second coil increases at a rate of 0.3 a/s, what is the magnitude of the induced emf in the first coil?

Explanation:

We have two which are wound around in the same form

The rate decrease of current in the first coils is

di₁/dt = —0.25A/s

Induced emf in the second coil is

ε₂ = 1.65 × 10^-3 V

A. Mutual inductance?

The induced emf in the second coil in terms of the mutual inductance and the rate of the change in current is given as

ε₂ = M•|di₁/dt|

Where M is mutual inductance

Making M subject of formula

M = ε₂ / di₁/dt

Then, M = 1.65 × 10^-3 / 0.25

M = 6.6 × 10^-3 H

B. If the second coil has a turn of 30turns

N₂ = 30turns

Also, current in the first coil is

i₁ = 1.25A

Flux Φ?

The average flux in the second coil is given as

Φ₂ = M•i₁ / N₂

Φ₂ = 6.6 × 10^-3 × 1.25 / 30

Φ₂ = 2.75 × 10^-4 Wb

C. The current in the second coil increase at rate of

di₂/dt = 0.3 A/s

Induce emf in coil 1 ε₁?

ε₁ = M•|di₂/dt|

ε₁ = 6.6 × 10^-3 × 0.3

ε₁ = 1.98 × 10^-3 V

b) phosphorus pentoxide

  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