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3 years ago

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In each of two coils the rate of change of the magnetic flux in a single loop is the same. The emf induced in coil 1, which has

228 loops, is 3.13 V. The emf induced in coil 2 is 4.16 V. How many loops does coil 2 have

1 answer:

Nat2105 [25]3 years ago

5 0

Answer:

303

Explanation:

We are given that

Emf in coil 1, $E_1=3.13 V$

Emf induced in coil 2, $E_2=4.16 V$

Number of loops in coil 1, $N_1=228$

We have to find the number of loops in coil 2.

Rat of change of magnetic flux in a single loop is same.

Let $\phi_1$ and $\phi_2$ be the magnetic flux in coil 1 and coil 2.

$\frac{d\phi_1}{dt}=\frac{d\phi_2}{dt}$

$\frac{V_2}{V_1}=\frac{N_2}{N_1}$

Using the formula

$\frac{E_2}{E_1}=\frac{N_2}{N_1}$

$\frac{4.16}{3.13}=\frac{N_2}{228}$

$N_2=\frac{4.16}{3.13}\times 228$

$N_2=303$

Hence, the number of loops in coil 2=303

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A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If

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Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

$\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s$

The induced emf in the shorter coil is calculated as;

$E = NA\frac{\delta B}{\delta t}$

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

$E = NA\frac{\delta B}{\delta t}$

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

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3 years ago

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Answer:

T=4.24 N.m

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Torque is equal to force for distance for sinus of the angle between the direction of the force and the distance, the distance between the mass and the pivot is 1 m, and to obtain the force that is the mass for the gravity in this case, we need to know the component that produces a torque in the pivot

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anothe way to do it is,

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3 years ago

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Answer:

Explanation:

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$V_2=0.264 m^3$

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Since it is a polytropic Process

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$W=57.48 kJ$

From Energy balance

$E_{in}-E_{out}=\Delta E_{system}$

Neglecting kinetic and Potential Energy change

$Q_{in}+W_{in}=change\ in\ Internal\ Energy$

Change in Internal Energy $\Delta U=u_2-u_1$

$\Delta U=mc_v(T_2-T_1)$

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$\Delta U=45.977 kJ$

$Q_{in}+57.48=45.977$

$Q_{in}=-11.50 kJ$

i.e. Heat is being removed

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3 years ago