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Alex Ar [27]
2 years ago
10

Magnetic field strength decreases as you get farther from the poles of the magnet. does the flux through the face of the coil ch

ange as you move the magnet? explain your answer.
Physics
1 answer:
brilliants [131]2 years ago
5 0

Magnetic field strength decreases as you get farther from the poles of the magnet.

<h3>Explanation</h3>

There are three ways to change the magnetic flux across a loop: Change the strength of the magnetic field across the surface (raise, reduce).

It doesn’t matter whether you move the coil while keeping the magnet fixed or the other way around if you want to change the surface area of the loop (raise by expanding the loop, decrease by reducing the loop).

The way they move in respect to one another determines how the magnetic flux across the coil changes. The direction in which the galvanometer needle deflects depends on which side of the coil confronts the magnet when you move it. The coil ends are labelled to show which goes to the red wire and which goes to the black wire.

To know more about magnetic flux visit:

brainly.com/question/16960581

#SPJ4

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LuckyWell [14K]

Answer:

B = 1058.4  N

Explanation:

Given that,

The volume of a metal block, V = 0.09 m³

The density of fluid, d = 1200 kg/m³

We need to find the buoyant force when it's Completely  immersed in brine. The formula for the buoyant force is given by :

B=\rho gV

g is acceleration due to gravity

B=1200\times 9.8\times 0.09\\\\B=1058.4\ N

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Answer:

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3 years ago
Read 2 more answers
A coil 3.95 cm radius, containing 520 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×1
Pie

Answer:

(a) E= 3.36×10−2 V +( 3.30×10−4 V/s3 )t3

(b) I=0.0085\ A

Explanation:

Given:

  • radius if the coil, r=0.0395\ m
  • no. of turns in the coil, n=520
  • variation of the magnetic field in the coil, B=(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4
  • resistor connected to the coil, R=560\ \Omega

(a)

we know, according to Faraday's Law:

emf=n.\frac{d\phi}{dt}

where:

d \phi= change in associated magnetic flux

\phi= B.A

where:

A= area enclosed by the coil

Here

A=\pi.r^2

A=\pi\times 0.0395^2

A=0.0049\ m^2

\therefore \phi=((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049

So, emf:

emf= 520\times \frac{d}{dt} [((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049]

emf= 520\times 0.0049\times \frac{d}{dt} [(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)]

emf= 2.548\times [0.012+(13.8\times 10^{-5})t^3)]

emf= 0.0306+3.516\times 10^{-4}\ t^3

(b)

Given:

t_0=5.25\ s

Now, emf at given time:

emf=4.7755\times 10^{-2}\ V

∴Current

I=\frac{emf}{R}

I=\frac{4.7755\times 10^{-2}}{560}

I=8.5\times 10^{-5} A

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i) A 100 W and 60 W bulb are joined in series and connected to the mains. Which bulb will glow brighter? Why? ii) A 100 W and a
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Answer:

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ii) 100 W

Explanation:

In each case, the bulb that dissipates the most power is the bulb that glows brighter.  Power is voltage times current (P = VI).  Using Ohm's law, we can rewrite this as P = I²R or P = V²/R.

Bulbs are rated at a certain power for a certain voltage.  P = V²/R, so the bulb with the lower resistance will have the higher power rating.  Therefore, the 100 W bulb has a lower resistance than the 60 W bulb.

i) They are in series, so they have the same current.  P = I²R, so the bulb with the higher resistance will glow brighter.  That's the 60 W bulb.

ii) They are in parallel, so they have the same voltage.  P = V²/R, so the bulb with the lower resistance will glow brighter.  That's the 100 W bulb.

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