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melisa1 [442]
3 years ago
12

Suppose the first coil of wire illustrated in the simulation is wound around the iron core 7 times. Suppose the second coil is w

ound around the core N2 = 13 times. Shortly after the switch has been closed, the flux through the second coil of wire changes from 2.1 Wb per turn to 2.7 Wb in each turn of wire over a time interval of 0.3 s. Find the magnitude of the emf induced in the coil.
Physics
2 answers:
Elena L [17]3 years ago
8 0

Answer:

emf=26\ V

Explanation:

Given:

no. of turns in the first coil, n_1=7

no. of turns in the second coil, n_2=13

the change in flux through the secondary coil, d\phi=2.7-2.1=0.6\ Wb

time taken for the change in flux, dt=0.3\ s

  • According to the Faraday's law there will be an emf induced in the coil associated with the rate of change in magnetic flux.

<u>This emf is mathematically given as:</u>

emf=n_2\times \frac{d\phi}{dt}

emf=13\times \frac{0.6}{0.3} (we take no. of turns of the second coil because the rate of change in flux is associated with the second coil)

emf=26\ V

SVETLANKA909090 [29]3 years ago
5 0

Answer:

Emf induced in the coli will be equal to 26 volt

Explanation:

We have given number of turns N = 13

It is given that magnetic flux changes from 2.1 Weber to 2.7 Weber in 0.3 sec

Change in flux d\Phi =2.7-2.1=0.6Weber

Change in time dt = 0.3 sec

We have to find the induced emf

Induced emf is equal to e=N\frac{d\Phi }{dt}=13\times \frac{0.6}{0.3}=26volt

So emf induced in the coil will be equal to 6 volt

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A smooth circular hoop with a radius of 0.800 m is placed flat on the floor. A 0.300-kg particle slides around the inside edge o
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Answer:

a)11.25 J

b)Number of revolution = 1

Explanation:

Given that

Radius ,r= 0.8 m

m= 0.3 kg

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final speed ,v= 5 m/s

a)

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Final kinetic energy

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b)

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Now by putting the values

V=\sqrt{0.8\times 10}

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KE=\dfrac{1}{2}mV^2

KE=\dfrac{1}{2}0.3\times 2.82^2

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Here 22.5 J is greater than 13.8 J.So the particle will complete only one revolution.

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4 0
3 years ago
the length of iron rod at 100 C is 300.36 cm and at 159 C is 300.54 cm.Calculate its length at 0 c and coefficient of linear exp
Ugo [173]

Answer:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

Explanation:

From the question given above, the following data were obtained:

Length (L₁) at 100 °C = 300.36 cm

Temperature 1 (θ₁) = 100 °C

Length (L₂) at 159 °C = 300.54 cm

Temperature 2 (θ₂) = 159 °C

Length (L₀) at 0 °C =?

Coefficient of linear expansion (α) =?

L₁ = L₀ (1 + θ₁α)

300.36 = L₀ (1 + 100α) ....(1)

L₂ = L₀ (1 + θ₂α)

300.54 = L₀ (1 + 159α) ..... (2)

Divide equation (2) by (1)

300.54 / 300.36 = L₀ (1 + 159α) / L₀ (1 + 100α)

1.0006 = (1 + 159α) / (1 + 100α)

Cross multiply

1.0006 (1 + 100α) = (1 + 159α)

1.0006 + 100.06α = 1 + 159α

Collect like terms

1.0006 – 1 = 159α – 100.06α

0.0006 = 58.94α

Divide both side by 58.94

α = 0.0006 / 58.94

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Substitute the value of α into anything of the equation to obtain L₀. Here we shall use equation (2).

300.54 = L₀ (1 + 159α)

α = 1.02×10¯⁵ C¯¹

300.54 = L₀ (1 + 159 ×1.02×10¯⁵)

300.54 = L₀ (1 + 0.0016218)

300.54 = L₀ (1.0016218)

Divide both side by 1.0016

L₀ = 300.54 / 1.0016

L₀ = 300.05 cm

Summary:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

6 0
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