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77julia77 [94]
3 years ago
9

A current I flows down a wire of radius a.

Physics
1 answer:
Helga [31]3 years ago
7 0

Answer:

(a) K = \frac{I}{2\pi a}

(b) J = \frac{I}{2\pi as}

Explanation:

(a) The surface current density of a conductor is the current flowing per unit length of the conductor.

                                   K = \frac{dI}{dL}

Considering a wire, the current is uniformly distributed over the circumferenece of the wire.

                                   dL = 2\pi r

The radius of the wire = a

                                    dL = 2\pi a

The surface current density K = \frac{I}{2\pi a}

(b) The current density is inversely proportional

                                     J \alpha  s^{-1}    

                                     J = \frac{k}{s}           ......(1)

k is the constant of proportionality

                                     I = \int\limits {J} \, dS

                                     I = J \int\limits \, dS     ........(2)

substituting (1) into (2)

                                     I = \frac{k}{s} \int\limits\, dS

                                     I = k \int\limits^a_0 \frac{1}{s}  {s} \, dS

                                     I = 2\pi k\int\limits\, dS

                                     I = 2\pi ka

                                     k = \frac{I}{2\pi a}

substitute J = \frac{k}{s}

                                     J = \frac{I}{2\pi as}

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Explanation:

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A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul
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Here in this case we can use work energy theorem

As per work energy theorem

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KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2

also we know that

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KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2

KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2

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If at 10m above the ground an object has 50J of Kinetic Energy, with 50J of Potential Energy. How high can the object travel?
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Hi there!

\large\boxed{\text{B) 20 meters}}

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K = Kinetic Energy (J)

E = Total Energy (J)

At 10m, the total amount of energy is equivalent to:

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100J = 5h

<u>h = 20 meters</u>

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