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Mnenie [13.5K]
1 year ago
13

a square loop whose sides are long is made with copper wire of radius , assuming resistivity of copper is . if a magnetic field

perpendicular to the loop is changing at a constant rate of , what is the current in the loop?
Physics
1 answer:
OLEGan [10]1 year ago
4 0

A square loop whose sides are long is made of copper wire of radius , given the resistivity of copper is . if the magnetic field perpendicular to the loop changes at a constant rate of I = 14.029 mA.

The basic characteristic of a substance that measures how effectively it resists an electric current is called electrical resistance. A material with low resistance is a material that easily conducts electric current. A Greek letter is often used to indicate resistivity. Electrical resistance is a basic property of a material that measures how strongly it resists an electric current. The SI unit for electrical resistance is the ohmmeter.

We use magnetic field as a tool to describe how the magnetic field is distributed in the space around and inside something of a magnetic nature. A material with low resistance is a material that easily conducts electric current. A Greek letter is often used to indicate resistivity. An ohmmeter is a unit of electrical resistance in the SI system.

Learn more about magnetic field here;

brainly.com/question/24397546

#SPJ4

The complete question is :

A square loop whose sides are 6.0-cm long is made with copper wire of radius 1.0 mm. If a magnetic field perpendicular to the loop is changing at a rate of 5.0 mT/s, what is the current in the loop?

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Answer:

Island arc

Explanation:

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3 years ago
(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th
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Answer:

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Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

V = -\dfrac{kQ}{r}

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(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

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3 years ago
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Part a.
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Because the model bridge weighs 50 N, the real bridge weighs
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Part b.
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3 years ago
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Let  us consider two bodies having masses m and m' respectively.

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From the above we see that F ∝ mm' and F\alpha \frac{1}{r^{2} }

Let the orbital radius of planet  A is r_{1}  = r and mass of planet is m_{1}.

Let the mass of central star is m .

Hence the gravitational force for planet A  is f_{1} =G \frac{m_{1}*m }{r^{2} }

For planet B the orbital radius  r_{2} =2r_{1} and mass m_{2} = 3 m_{1}

Hence the gravitational force f_{2} =G\frac{m m_{2} }{r^{2} }

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3 0
3 years ago
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Explanation:

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Speed of a plane a is 900 km/h and that of plane b is 50 km/h. So, we can say that plane a is moving faster.

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3 years ago
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