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Doss [256]
3 years ago
10

a circular loop is hanging on the wall. it has a radius of 33.3 cm and is comprised of 12 coils. there is a magnetic field perpe

ndicular to the wall, coming out of the wall, that grows from 1.5 T to 7.5 T. the coil has a resistance of 3.75 ohms, what is the magnitude and direction of the induced current in the coil?
Physics
1 answer:
Vika [28.1K]3 years ago
5 0

Answer:

I = 2.19A, anticlockwise direction.

Explanation:

Given r = 33cm = 0.33m, N = 12, ΔB = 7.5 - 1.5 = 6.0T, Δt = 3s, R = 3.75Ω

By Faraday's law of electromagnetic induction when there is a change in flux in a coil or loop, an emf is induced in the coil or loop which is proportional to the time rate of change of the magnetic flux through the loop.

The emf E is related to the flux by the formula

E = – NdФ/dt

Where N = number of turns in the coil, Ф = magnetic flux through the loop = BA, B = magnetic field strength, A = Area

In this problem the strength of the magnetic field changes. As a result the flux too changes and an emf is induced in the coil.

So

ΔФ = ΔB×A = ΔB×πr² = 6×π×0.33² = 2.05Wb

E = -NΔФ/Δt = 12×2.05/3 = 8.2V

I = E/R = 8.2/ 3.75 = 2.19A

The direction of the current can be found by pointing the thumb of your right hand in the direction of the magnetic field and curling the remaining fingers around this direction. The direction of the curl of these fingers give the direction of current which in this case is anticlockwise.

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The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

\dfrac{1}{2} \times 166 \times v^2 = 17457.0912

v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2)  }}

Which gives;

\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66)  } \approx 8.1

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

Time = \dfrac{Distance}{Velocity}

\mathrm{The \ time \ the\ Lone \  Ranger \  and  \ Tonto \  have,  \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

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Explanation:

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