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OLEGan [10]
2 years ago
7

A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq

ual to that of the frame. A uniform magnetic field is turned on perpendicular to the plane of the coil. This field changes at a constant rate from 20 to 80 mT in a time of 20 ms. What is the magnitude of the induced emf in the coil at the instant the magnetic field has a magnitude of 50 mT?
Physics
1 answer:
MAVERICK [17]2 years ago
9 0

Answer:

Approximately 18 volts when the magnetic field strength increases from \rm 20\; mT to \rm 80\;mT at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF \epsilon that a changing magnetic flux induces in a coil is:

\displaystyle \epsilon = N \cdot \frac{d\phi}{dt},

where

  • N is the number of turns in the coil, and
  • \displaystyle \frac{d\phi}{dt} is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux \phi is equal to

\phi = B \cdot A\cdot \cos{\theta},

where

  • B is the magnetic field strength at the coil, and
  • A\cdot \cos{\theta} is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so \theta = 0 and A\cdot \cos{\theta} = A. The area of this circular coil is equal to \pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}.

A\cdot \cos{\theta} = A doesn't change, so the rate of change in the magnetic flux \phi through the coil depends only on the rate of change in the magnetic field strength B. The size of the magnetic field at the instant that B = \rm 50\; mT will not matter as long as the rate of change in B is constant.

\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}.

As a result,

\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V.

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Answer:

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Explanation:

Given:

  • test charge +q
  • distance of the test charge from +Q, r
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<u>Force on the test charge due to +Q:</u>

F_1=k.\frac{Q.q}{r^2}

<u>Force on the test charge due to +Q:</u>

F_2=k.\frac{2Q.q}{(2r)^2}

F_2=k.\frac{Q.q}{2r^2}

Since all the charges are positive here, so they will try to repel the test charge away. And the force due to charge +Q will be greater so initially the test charge will move rightwards away from the +Q charge.

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2 years ago
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Answer:

ΔU=0.8834 Btu

Explanation:

Given data

Area of piston A=40 in²

The weight W=100 lbf

Atmospheric Pressure P=14.7 lbf/in²

Work added E=3 Btu

The change in elevation Δh=1 ft =12 inch

To find

Change in internal energy of the gas ΔU

Solution

For Piston

ΔPE=| W+(P×A)×Δh |

ΔPE=| 100+(14.7×40)×12 |

ΔPE=8256 lbf.in

ΔPE=8256×0.000107

ΔPE=0.8834 Btu

From law of conservation of energy then ,the charging in the potential energy of the piston is made by exerting force by gas

Wgas= -ΔPE

Wgas= -0.8834 Btu

For the gas as a system and by applying first law of thermodynamics

Q-W=ΔU

0-(-0.8834 Btu)=ΔU

So

ΔU=0.8834 Btu

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2 years ago
At a baseball game, Bill observes from the bleachers Jack throwing a ball toward home at 15 km/s while Kevin runs toward home fr
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Answer:bill 5 m/s. Jack:10 m/s

Explanation:

Cuz I took it

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An object is released from rest at a height h. During the final second of its fall, it traverses a distance of 38m. Determine th
nadezda [96]
Gravity is 9.8 m/s² means means every second distance travelled
increases by the distance in the previous second plus an extra 9.8m
during last second it fell 38m
previous second dist = 38 - 9.8m = 28.2
previous second = 28.2 - 9.8m = 18.4m
distance left = 18.4 - 9.8m = 8.6m
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Two astronauts, each having a mass of 75.0 kg, are connected by a 10.0-m rope of negligible mass. They are isolated in space, or
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Answer:

a) 3750 kgm²/s

b) 1875 J

c) 3750 kgm²/s

d) 10 m/s

e) 7500 J

f) 5625 J

Explanation:

a)

d = distance between the two astronauts = 10 m

m = mass of each astronaut = 75 kg

v = speed of each astronaut = = 5 m/s

r = distance of each astronaut from center of mass = (0.5) d = (0.5) (10) = 5 m

L_{s} = Angular momentum of the system

Angular momentum of the system is given as

L_{si} = 2 m v r

L_{si} = 2 (75) (5) (5)

L_{si} = 3750 kgm²/s

b)

Rotational energy of the system is given as

R_{si} = 2 (0.5) m v²

R_{si} = 2 (0.5) (75) (5)² = 1875 J

c)

L_{sf} = Angular momentum of the system after pulling

Using conservation of angular momentum,

L_{sf} = L_{si}

L_{sf} = 3750 kgm²/s

d)

r' = new distance of each astronaut from center of mass = (0.5) r = (0.5) (5) = 2.5 m

v' = new speed of each astronaut

Angular momentum of the system after pulling is given as

L_{sf} = 2 m v' r'

3750 = 2 (75) (2.5) v'

v' = 10 m/s

e)

Rotational energy of the system after pulling the rope is given as

R_{sf} = 2 (0.5) m v'²

R_{sf} = 2 (0.5) (75) (10)² = 7500 J

f)

W = work done by astronaut in pulling the rope

Work done is given as

W = R_{sf} - R_{si}

W = 7500 - 1875

W = 5625 J

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2 years ago
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