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lyudmila [28]
3 years ago
5

A single-turn circular loop of wire of radius 45 mm lies in a plane perpendicular to a spatially uniform magnetic field. During

a 0.10 s time interval, the magnitude of the field increases uniformly from 250 to 350 mT. (a) Determine the emf induced in the loop (in V). (Enter the magnitude.) V (b) If the magnetic field is directed out of the page, what is the direction of the current induced in the loop
Physics
1 answer:
Lina20 [59]3 years ago
7 0

Answer:

Magnitude of induced emf is 0.00635 V

Explanation:

Radius of circular loop r = 45 mm = 0.045 m

Area of circular loop A=\pi r^2

A=3.14\times 0.045^2=0.00635m^2

Magnetic field is increases from 250 mT to 350 mT

Therefore change in magnetic field dB=250-350=100mT

Emf induced is given by

e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}

e=-0.00635\times \frac{100\times 10^{-3}}{0.10}=-0.00635V

Magnitude of induced emf is equal to 0.00635 V

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Two point charges lie on the x axis. A charge of + 2.30 pC is at the origin, and a charge of − 4.50 pC is at x=−11.0cm.
Radda [10]

r₁ = distance of point A from charge q₁ = 0.13 m

r₂ = distance of point A from charge q₂ = 0.24 m

r₃ = distance of point A from charge q₃ = 0.13 m

Electric field by charge q₁ at A is given as

E₁ = k q₁ /r₁² = (9 x 10⁹) (2.30 x 10⁻¹²)/(0.13)² = 1.225 N/C     towards right

Electric field by charge q₂ at A is given as

E₂ = k q₂ /r₂² = (9 x 10⁹) (4.50 x 10⁻¹²)/(0.24)² = 0.703 N/C    towards left

Since the electric field in left direction is smaller, hence the electric field by the third charge must be in left direction

Electric field at A will be zero when

E₁ = E₂ + E₃

1.225 = 0.703 + E₃

E₃ = 0.522 N/C

Electric field by charge "q₃" is given as

E₃ = k q₃ /r₃²

0.522 = (9 x 10⁹) q₃/(0.13)²

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3 years ago
How to delete question​
Marrrta [24]
Oh your from the other question you made I just saw it LOL.
But heres the answer click the 3 dots on the question you made or you can ask a Moderator or Administrator to remove your question with a reason.
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3 years ago
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If the temperature is held constant during this process and the final pressure is 683 torrtorr , what is the volume of the bulb
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Answer:

Explanation:

Let the volume of the unknown bulb = X L

The volume of the system , after opening valve = (X + 0.72 L )

Use Boyles law gas equation,

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Given:

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1 atm = 760 mmHg(torr)

683 mmHg = 683/760

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1.2X = 0.8987*(X + 0.720)

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