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Elanso [62]
3 years ago
15

A planar loop consisting of seven turns of wire, each of which encloses 200 cm2, is oriented perpendicularly to a magnetic field

that increases uniformly in magnitude from 14 × 10–3 T to 38 × 10–3 T in a time of 8.0 × 10–3 s. What is the resulting induced current in the coil if the resistance of the coil is 5.0 Ω?
Physics
2 answers:
evablogger [386]3 years ago
6 0

Given Information:  

Area of loop = A = 200 cm² = 0.0200 m²

Change in time = Δt = 8x10⁻³ seconds

Change in magnetic field = ΔB = (38x10⁻³ - 14x10⁻³) T  

Number of turns = N = 7

Resistance of coil = R = 5 Ω

Required Information:  

Induced current = I = ?  

Answer:

Induced current = 0.084 A

Explanation:

From the Faraday's law the induced EMF ξ in the coil is given by

ξ = -NΔΦ/Δt

Where ΔΦ is the change in flux

ΔΦ = ΔBA

Where A is the area of the planer loop

ΔΦ = (38x10⁻³ - 14x10⁻³)*0.0200

ΔΦ = 0.00048

So the induced emf becomes

ξ = -NΔΦ/Δt

ξ = (-7*0.00048)/8x10⁻³

ξ = -0.42 V

The negative sign indicates that the induced emf opposes the change that produced it in the first place.

Finally, we can now find the induced current using Ohm's law

I = ξ/R

I = 0.42/5

I = 0.084 A

Therefore, the resulting induced current in the coil is 0.084 A

PtichkaEL [24]3 years ago
3 0

Answer:

The induced current is 0.084 A

Explanation:

the area given by the exercise is

A = 200 cm^2 = 200x10^-4 m^2

R = 5 Ω

N = 7 turns

The formula of the emf induced according to Faraday's law is equal to:

ε = (-N * dφ)/dt = (N*(b2-b1)*A)/dt

Replacing values:

ε = (7*(38 - 14) * (200x10^-4))/8x10^-3 = 0.42 V

the induced current is equal to:

I = ε /R = 0.42/5 = 0.084 A

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a tank circuit contains a capacitor and an inductor that produce 30 of reactance at the resonant frequency. the inductor has a q
Romashka-Z-Leto [24]

The total circuit current at the resonant frequency is 0.61 amps

What is a LC Circuit?

  • A capacitor and an inductor, denoted by the letters "C" and "L," respectively, make up an LC circuit, also referred to as a tank circuit, a tuned circuit, or a resonant circuit.
  • These circuits are used to create signals at particular frequencies or to receive signals from more complicated signals at particular frequencies.

Q =15 = (wL)/R

wL = 30 ohms = Xl

R = 2 ohms

Zs = R + jXl = 2 +j30 ohms where Zs is the series LR impedance

| Zs | = 30.07 <86.2° ohms

Xc = 1/(wC) = 30 ohms

The impedance of the LC circuit is found from:

Zp = (Zs)(-jXc)/( Zs -jXc)

Zp = (2+j30)(-j30)/(2 + j30-j30) = (900 -j60)2 = 450 -j30 = 451 < -3.81°

I capacitor = 277/-j30 = j9.23 amps

I Zs = 277/(2 +j30) = (554 - j8,310)/904 = 0.61 - j9.19 amps

I net = I cap + I Zs = 0.61 + j0.04 amps = 0.61 < 3.75° amps

Hence, the total circuit current at the resonant frequency is 0.61 amps

To learn more about LC Circuit from the given link

brainly.com/question/29383434

#SPJ4

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