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AVprozaik [17]
3 years ago
5

A circuit contains a battery, a switch, an inductor, and a resistor connected in series. Initially, the switch is open.

Physics
1 answer:
tia_tia [17]3 years ago
8 0

Answer:

a long time after the switch is closed

Explanation:

Energy stored in an inductor is given by

E = 1/2 L i²

A). Before the switch is closed the current in the circuit is zero so energy in inductor will be zero.

B). Inductor opposes the flow of current. It takes some time for current to reach its maximum value. So immedialtly after closing the switch the current  and energy in inductor will still be zero.

C). After closing the switch it takes some time for current to reach its maximum value as inductor opposes the flow of current. so after a long time the current in inductor will reach its maximum value and hence the energy stored will be maximum.

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a horizontal force of 50N acts on a body of mass 20KG initially at rest and displaces it by striking it along the entire 20m pat
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all the forces occur in pairs that if one object exerts a force on another object, then the second object exerts an equal and opposite reaction force on the first.

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How does volcanic ash in earth's atmosphere affect solar radiation?
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A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne
kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

Emf = -N\frac{\Delta \phi}{\Delta t}

Where N is the number of turns

\Delta \phi is the change in magnetic  flux

\Delta t is the time interval

The change in magnetic flux is given by

\Delta \phi = \phi _{f} - \phi _{i}

Where \phi _{f} is the final magnetic flux

and \phi _{i} is the initial magnetic flux

Magnetic flux is given by the formula

\phi = BAcos(\theta)

Where B is the magnetic field

A is the area

and \theta is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, \theta = 0^{o}

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

A = \pi r^{2}

∴ A = \pi (0.15)^{2}

A = 0.0225\pi

∴\phi_{i}  = (1.5)(0.0225\pi)cos(0^{o} )

\phi_{i}  = (1.5)(0.0225\pi)

( NOTE: cos (0^{o}) = 1 )

\phi_{i}  = 0.03375\pi Wb

For \phi_{f}

The field pointed upwards, that is \theta = 90^{o}. Since cos (90^{o}) = 0

Then

\phi_{f} = 0

Hence,

\Delta \phi = 0- 0.03375\pi

\Delta \phi = - 0.03375\pi

From the question

\Delta t = 2.8 ms = 2.8 \times 10^{-3} s

Here, N = 1

Hence,

Emf = -N\frac{\Delta \phi}{\Delta t} becomes

Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }

Emf = 37.9 V

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0
3 years ago
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