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Elodia [21]
3 years ago
7

A 0.130 m radius, 485-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a

uniform magnetic field. (This is 60 rev/s.) Find the magnetic field strength needed to induce an average emf of 10,000 V.
Physics
2 answers:
Lady_Fox [76]3 years ago
5 0

Answer:

The magnetic field strength needed is 1.619 T

Explanation:

Given;

Number of turns, N = 485-turn

Radius of coil, r = 0.130 m

time of revolution, t =  4.17 ms = 0.00417 s

average induced emf, V = 10,000 V.

Average induced emf is given as;

V = -ΔФ/Δt

where;

ΔФ is change in flux

Δt is change in time

ΔФ = -NBA(Cos \theta_f - Cos \theta_i)

where;

N is the number of turns

B is the magnetic field strength

A is the area of the coil = πr²

θ is the angle of inclination of the coil and the magnetic field,

\theta_f = 90^o\\\theta_i = 0^o

V = NBACos0/t

V = NBA/t

B = (Vt)/NA

B = (10,000 x 0.00417) / (485 x π x 0.13²)

B =1.619 T

Thus, the magnetic field strength needed is 1.619 T

Cloud [144]3 years ago
3 0

Answer:

B = 1.619 T

Explanation:

We are given;

Radius of coil, r = 0.13 m

Number of turns; N = 485-turn

Time of revolution, t = 4.17 ms = 0.00417 s

Average induced emf = 10,000

The equation for the for the induced emf is given by;

EMF = -N(ΔФ/Δt)

where;

N is number of turns of coil

ΔФ is change in magnetic flux

Δt is change in time interval

The change in magnetic flux is given by the formula ;

ΔФ = BAcosθ

where;

B is the magnetic field strength

A is the area of the coil = πr²

θ is the angle between the normal to the plane and the magnetic field,

Now, since the coil rotates through one fourth of a revolution, the value of θ changes from 0° to 90°

The value of Δcosθ is found as;

Δcosθ = cos90 - cos0 = 0 - 1 = - 1

Thus,

ΔФ = BAcosθ = -BA

And so, EMF = -N(-BA/Δt) = NBA/Δt

Making B the subject, we have;

B = ((EMF)Δt)/NA

A = πr² = π x 0.13² = 0.0531 m²

Thus, B = ((10000•0.00417)/(485•0.0531) = 1.619 T

V = NBA/t

B = (Vt)/NA

B = (10,000 x 0.00417) / (485 x π x 0.13²)

B =1.619 T

Thus, the magnetic field strength needed is 1.619 T

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\Delta L=\Delta L_b+\Delta L_s\\\Rightarrow \Delta L=L_{0b}\alpha_b\Delta T+L_{0s}\alpha_b\Delta T\\\Rightarrow \Delta T=\frac{\Delta L}{L_{0b}\alpha_b+L_{0s}\alpha_b}\\\Rightarrow \Delta T=\frac{0.003}{0.31\times 18\times 10^{-6}+11\times 10^{-6}\times 11}\\\Rightarrow \Delta T=23.7\ ^{\circ}C

\Delta T=23.7\\\Rightarrow T_f-T_i=23.7\\\Rightarrow T_f=23.7+T_i\\\Rightarrow T_f=23.7+20\\\Rightarrow T_f=43.7\ ^{\circ}C

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(a) is correct option

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Weight of nitrogen = 20 g

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We need to calculate the internal energy

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Q=\Delta U+nRT

Put the value into the formula

4480=\Delta U+\dfrac{20}{28}\times8.314\times77

\Delta U=4480-\dfrac{20}{28}\times8.314\times77

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We need to calculate the number of molecules in 20 g N₂

Using formula of number of molecules

N=n\times \text{Avogadro number}

Put the value into the formula

N=\dfrac{20}{28}\times6.02\times10^{23}

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We need to calculate the energy

Using formula of energy

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Put the value into the formula

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