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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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Step 1) They will start to form liquids, like condensation.
Step 2) As liquids cool, they will turn into solids.
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3 years ago
Which of the following is an example of Newton's Third Law?* O A stack of pennies will not move unless you flick them over. O Fa
Morgarella [4.7K]

Answer:

A ball hits the ground and the ground pushes up on it

Explanation:

Newton's third law basically states that for every action, there's a reaction.

a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.

Hope this Helps!!! :)

5 0
3 years ago
The lithosphere consists of _______. select one:
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3 0
3 years ago
If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

s_x=v_x_0 t

t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s

Now for the vertical distance

vy_o=0

than the equation of motion becomes

s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2

Now using this acceleration the value of electric field is calculated as

E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0
3 years ago
Please help soon!!!!
anyanavicka [17]
The Answer: Parallel
6 0
3 years ago
Read 2 more answers
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