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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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Please help with these questions as well! I need urgent help! I will give brainliest! God bless!
Radda [10]

6.  Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.  

F = ma

F = (15.3)(-9.8)

F = -150N

Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object.  So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.


7.  Same idea as question 2.  

First determine the weight of the object:

F = ma

F = (30)(-9.8)

F = -294N

The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.

-294 + 150N + x = 0

x = 144N  

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10.  First find the force of block B to the right due to its acceleration:

F = ma

F = (24)(0.5)

F = 12N

So block B is moving 12N to the right relative to block A due to block A's movement to the left.  However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A.  The force that is causing block B to experience the lower relative force to the right is because of the friction.  To find the friction:

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5 0
3 years ago
Underground water is being pumped into a pool whose cross section is 3 m x 4 m while water is discharged through a 0.076m-diamet
Svetllana [295]
Given:

Area of pool = 3m×4m
Diameter of orifice = 0.076m
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Accumulation velocity = 1.5cm/min

Required:

Inlet flowrate

Solution:

The problem can be solved by this general formula.

Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

First, we need to convert the units of the accumulation velocity into m/s to be consistent.

Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s

We then calculate the area of the pool and the area of the orifice by:

Area of pool = 3 × 4 m²
Area of pool = 12m²

Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²

Since we have all we need, we plug in the values to the general equation earlier

Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²

Transposing terms,

Inlet flowrate = 0.316 m³/s
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MA_775_DIABLO [31]

Answer:

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Explanation:

The sun radiates energy to the earth to make it warmer near the equator.

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