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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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6 0
3 years ago
Read 2 more answers
A. What is the RMS speed of Helium atoms when the temperature of the Helium gas is 343.0 K? (Possibly useful constants: the atom
kkurt [141]

Answer:

(a) 1462.38 m/s

(b) 2068.13 m/s

Explanation:

(a)

The Kinetic energy of the atom can be given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 10⁻²³ J/k

K.E = Kinetic Energy of atoms = 343 K

T = absolute temperature of atoms

The K.E is also given as:

K.E = (1/2)mv²

Comparing both equations:

(1/2)mv² = (3/2)KT

v² = 3KT/m

v = √[3KT/m]

where,

m = mass of Helium = (4 A.M.U)(1.66 X 10⁻²⁷ kg/ A.M.U) = 6.64 x 10⁻²⁷ kg

v = RMS Speed of Helium Atoms = ?

Therefore,

v = √[(3)(1.38 x 10⁻²³ J/K)(343 K)/(6.64 x 10⁻²⁷ kg)]

<u>v = 1462.38 m/s</u>

(b)

For double temperature:

T = 2 x 343 K = 686 K

all other data remains same:

v = √[(3)(1.38 x 10⁻²³ J/K)(686 K)/(6.64 x 10⁻²⁷ kg)]

<u>v = 2068.13 m/s</u>

8 0
3 years ago
a 1210 kg roller coaster car is moving 6.33 m/s. as it approaches the station, brakes slow it down to 2.38 m/s over a distance o
Stels [109]

Answer:

4960 N

Explanation:

First, find the acceleration.

Given:

v₀ = 6.33 m/s

v = 2.38 m/s

Δx = 4.20 m

Find: a

v² = v₀² + 2aΔx

(2.38 m/s)² = (6.33 m/s)² + 2a (4.20 m)

a = -4.10 m/s²

Next, find the force.

F = ma

F = (1210 kg) (-4.10 m/s²)

F = -4960 N

The magnitude of the force is 4960 N.

4 0
3 years ago
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