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anastassius [24]
9 months ago
7

15) a current carrying loop of wire lies flat on a table top. when viewed from above, the current moves around the loop in a cou

nterclockwise sense. for points inside the loop, the magnetic field caused by this current a) circles the loop in a clockwise direction. b) circles the loop in a counterclockwise direction. c) points straight up. d) points straight down. e) is zero.
Physics
1 answer:
Maksim231197 [3]9 months ago
6 0

The correct answer is option C.

Maxwell right hand thumb rule: if we hold a current carrying conductor in our right hand such that the thumb indicates the direction of current then the curling of fingers shows the direction of magnetic field around the conductor.

Here the current is counterclockwise so by use of Maxwell right hand thumb rule the direction of magnetic field is straight up.

Using Fleming right hand rule that States that if the fore-finger, middle finger and the thumb of left hand are stretched mutually perpendicular to each other, such that fore-finger points in the direction of magnetic field, the middle finger points in the direction of the motion of positive charge, then the thumb points to the direction of the force.

So, the current, when viewed from above, seems to move around loop in counterclockwise direction points straight up is because of Maxwell's Thumb Rule.

To know more about magnet, refer: brainly.com/question/17143116

#SPJ4

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Answer:

We can conclude 2.3x10^7 J was converted to thermal energy

Explanation:

<u>Energy Conservation </u>

According to the law of conservation of energy, the total energy of an isolated system must be constant. If some kind of energy is 'lost', we know it was transformed into another type.

Let's check the conditions of the problem. The ride vertically drops a distance of 71.6 m starting from rest, and at the bottom of the drop, its speed is 10 m/s. Knowing the mass of the cart plus passengers is 3.5X10^4 kg, we compute the total energy at the top of the drop.

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\displaystyle E=m.g.h=(35000)(9.8)(71.6)=2.5X10^7\ J

Now we compute the 'total' energy at the bottom (quoted because we know there is some mechanical energy loss in the drop)

\displaystyle E'=m.g.h'+\frac{mv'^2}{2}

This time h'=0 and v=10 m/s, thus

\displaystyle E'=\frac{(35000)10^2}{2}=1.8x10^6\ J

The mechanical energy at the top and the bottom are not the same, thus we can know part of it was converted to heat or thermal energy. We compute the difference

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We can conclude 2.3x10^7 J was converted to thermal energy

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