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Rom4ik [11]
3 years ago
7

The magnetic coils of a tokamak fusion reactor are in the shape of a toroid having an inner radius of 0.700 m and an outer radiu

s of 1.30 m. The toroid has 900 turns of largediameter wire, each of which carries a current of 14.0 kA. Find the magnitude of the magnetic field inside the toroid along (a) the inner radius and (b) the outer radius.
Physics
1 answer:
iragen [17]3 years ago
5 0

Answer:

(a) 11.3 T

(b) 6.09 T

Explanation:

Current, I = 14 kA = 14000 A

number of turns, N = 900

inner radius, r = 0.7 m

outer radius, R = 1.3 m

The magnetic field due to a circular coil is given by

B = \frac{\mu o}{4\pi}\times \frac{2 N\pi I}{R}

(a) The magnetic field due to the inner radius is

B = 10^{-7}\times \frac{2\times 900\times 3.14\times 14000}{0.7}\\\\B = 11.3 T

(b) The magnetic field due to the outer radius is

B = 10^{-7}\times \frac{2\times 900\times 3.14\times 14000}{1.3}\\\\B = 6.09 T

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What happens if the breakdown voltage is exceeded.
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A ball is thrown horizontally from the top of a 60 m building and lands 100 m from the base of the building. How long is the bal
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Answer:

The ball is in the air for 3.5 seconds

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The vertical component of the final velocity is 34.3 m/s downward

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

Explanation:

A ball is thrown horizontally

That means the vertical component of the initial velocity u_{y}=0

The initial velocity is the horizontal component u_{x}

The ball is thrown from the top of a 60 m

That means the vertical displacement component y = 60 m

→ y = u_{y} t + \frac{1}{2} gt²

where g is the acceleration of gravity and t is the time

y = -60 m , g = -9.8 m/s² , u_{y}=0

Substitute these values in the rule

→ -60 = 0 + \frac{1}{2} (-9.8)t²

→ -60 = -4.9t²

Divide both sides by -4.9

→ 12.2449 = t²

Take √ for both sides

∴ t = 3.5 seconds

* <em>The ball is in the air for 3.5 seconds </em>

The initial velocity is the horizontal component u_{x}

The ball lands 100 meter from the base of the building

That means the horizontal displacement x = 100 m

→ x = u_{x} t

→ t = 3.5 s , x = 100 m

Substitute these values in the rule

→ 100 = u_{x} (3.5)

Divide both sides by 3.5

→ u_{x} = 28.57 m/s

<em>The initial horizontal component of velocity is 28.6 m/s</em>

The vertical component of the final velocity is v_{y}

→ v_{y} = u_{y} + gt

→ u_{y} = 0 , g = -9.8 m/s² , t = 3.5 s

Substitute these values in the rule

→ v_{y} = 0 + (-9.8)(3.5)

→ v_{y} = -34.3 m/s

<em>The vertical component of the final velocity is 34.3 m/s downward</em>

The final velocity v is the resultant vector of  v_{x} and v_{y}

→ Its magnetude is v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}

→ Its direction tan^{-1}\frac{v_{y}}{v_{x}}

→ v_{y} = 28.6 , v_{y} = -34.3

Substitute this values in the rules above

→ v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66

→ Its direction tan^{-1}\frac{-34.3}{28.6}=-50.18

The negative sign means the direction is below the horizontal

<em>The final velocity is 44.7 m/s in the direction 50.2° below the horizontal</em>

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