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Nataly [62]
2 years ago
8

Two concentric, coplanar, circular loops of wire of different diameter carry currents in the same direction. Describe the nature

of the force exerted on the inner loop by the outer loop and on the outer loop by the inner loop.
Physics
1 answer:
Ierofanga [76]2 years ago
4 0

Answer:

Both will be attractive in nature.

Explanation:

In the given case, the direction of the magnetic field is same in both loops as the direction of the current is same in both loops. When two parallel straight wire carrying current in the same direction are brought close to each other, the force between them is attractive in nature. In the same way,  when two coplanar, circular and concentric loops of wire are carrying current in the same direction, the force between them is attractive in nature. It can be checked by using right hand thumb rule.

Check out other explanations.

brainly.com/question/15555539

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¿Qué tipo de nutrientes proporcionan energía a nuestro organismo?
kow [346]

Answer:

The correct answer is: macronutrients.

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7 0
3 years ago
Two large, parallel, nonconducting sheets of positive charge face each other. What is at points (a) to the left of the sheets, (
alexira [117]

Answer:

a)The electric Field will be zero at the point between the sheets

b)E_1=\dfrac{\sigma}{\epsilon_0}

c)E_2=\dfrac{\sigma}{\epsilon_0}

Explanation:

Let \sigma be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field  so the Electric Flux due to it is zero.

Now using Gauss law we have, E be the electric Field at the distance r from the sheet then

E\times 2A=\dfrac{\sigma A}{\epsilon_0}\\E=\dfrac{\sigma}{2\epsilon_0}

The Field will be away from the sheet and perpendicular to it.

a) The Electric Field between them

E_1=\dfrac{\sigma}{2\epsilon_0}-\dfrac{\sigma}{2\epsilon_0}\\=0

b)The Electric Field to the right of the sheets

E_1=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}

c)The Electric Field to the left of the sheets

E_2=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}

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