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xz_007 [3.2K]
3 years ago
14

In a region of space, a magnetic field points in the +x-direction (toward the right). Its magnitude varies with position accordi

ng to the formula Bx=B0+bx, where B0 and b are positive constants, for x≥0. A flat coil of area A moves with uniform speed v from right to left with the plane of its area always perpendicular to this field. Part A What is the emf induced in this coil while it is to the right of the origin?
Physics
1 answer:
otez555 [7]3 years ago
3 0

Answer:

E=Abv

Explanation:

We are given that

B_x=B_0+bx

For x\geq 0

Area of coil=A

Speed=v

We have to find the emf induced in the coil while it is to the right of the origin.

We know that induced emf

E=-\frac{d\phi}{dt}

\phi=BA

E=-\frac{d(A(B_0+bx)}{dt}

E=-A(b)\frac{dx}{dt}

We know that

v=-\frac{dx}{dt}

Substitute the value

E=Abv

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This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle?  Assume the specific heat ratio is 1.2.

Answer:

- stagnation pressure is 274.993 Mpa

- the stagnation temperature Tt is 4048 K

Explanation:

Given the data in the question;

To determine the stagnation pressure and temperature in the exit plane of the nozzle;

we us the expression;

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)

where Pt is stagnant pressure = ?

P is static pressure = 4 MPa = 4 × 10⁶ Pa  

Tt is stagnation temperature = ?

T is the static temperature  = 2000 K

γ is ratio of specific heats = 1.2

M is Mach number M = 3.2

we substitute

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)

Pt = 4 × 10⁶ × 68.7484

Pt = 274.993 × 10⁶ Pa

Pt = 274.993 Mpa

Therefore stagnation pressure is 274.993 Mpa

Now, to get our stagnation Temperature

Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa =  ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 =  Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

Tt⁶ = 4.3998976 × 10²¹

Tt = ⁶√(4.3998976 × 10²¹)

Tt = 4047.999 ≈ 4048 K

Therefore, the stagnation temperature Tt is 4048 K

6 0
3 years ago
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