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pav-90 [236]
3 years ago
8

A toroid having a square cross section, 4.76 cm on a side, and an inner radius of 22.0 cm has 740 turns and carries a current of

1.01 A. (It is made up of a square solenoid instead of round one bent into a doughnut shape.) What is the magnitude of the magnetic field inside the toroid at (a) the inner radius and (b) the outer radius?
Physics
1 answer:
Stels [109]3 years ago
3 0

Answer:

a. \beta_{in}=6.79x10^{-4}T

b. \beta_{out}=5.586x10^{-4}T

Explanation:

Given:

N=740, I=1.01A, u_o=4\pi*10^{-7} T*m/A.

The magnetic field can be find using the Gauss law in the equation of field magnetic at the inner radius and the outer radius knowing the inner radius and the outer radius are:

r_{in}=22.0cm*\frac{1m}{100cm}=0.22m

r_{out}=r_{in}+4.76=0.22+0.0476=0.2676m

a.

Magnetic field inner

\beta_{in}=\frac{u_o*I*N}{2\pi*r}

\beta_{in}=\frac{4\pi*10^{-7}T*m/A*1.01A*740}{2\pi*0.22m}

\beta_{in}=6.79x10^{-4}T

b.

Magnetic field outer

\beta_{out}=\frac{u_o*I*N}{2\pi*r}

\beta_{out}=\frac{4\pi*10^{-7}T*m/A*1.01A*740}{2\pi*0.2676m}

\beta_{out}=5.586x10^{-4}T

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7 0
2 years ago
A 3.35 kg object initially moving in the positive x direction with a velocity of 4.90 m s collides with and sticks to a 1.88 kg
ahrayia [7]

Answer:

The final components of velocity of the composite object is 3.33 m/s.

Explanation:

Given;

mass of the first object, m₁ = 3.35 kg

initial velocity of the first object, u₁ = 4.90 m/s in positive x-direction

mass of the second object, m₂ = 1.88 kg

initial velocity of the second object, u₂ = 3.12 m/s in negative y-direction

initial momentum of the first object, P₁ = 3.35 x 4.9 = 16.415 kgm/s

initial momentum of the second object, P₂ = 1.88 x 3.12 = 5.8656 kgm/s

The resultant velocity of the two objects is given by;

R² = 16.415² + 5.8656²

R² = 303.858

R = √303.858

R = 17.432 kgm/s

Apply the principle of conservation of linear momentum for inelastic collision;

total initial momentum before = total final momentum after collision

P₁(x) + P₂(y) = Pf

R = Pf

R = v(m₁ + m₂)

17.432 = v(m₁ + m₂)

where;

v is the final components of velocity of the composite object

v = \frac{17.432}{m_1 + m_2} \\\\v = \frac{17.432}{3.35+1.88} \\\\v = 3.33 \ m/s

Therefore, the final components of velocity of the composite object is 3.33 m/s.

8 0
3 years ago
Suppose a 4.0-kg projectile is launched vertically with a speed of 8.0 m/s. What is the maximum height the projectile reaches?
eduard

Answer:

h = 3.3 m (Look at the explanation below, please)

Explanation:

This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy

Kinetic energy = \frac{1}{2}mv^{2}

Plug in the numbers = \frac{1}{2}(4.0)(8^{2})

Solve = 2(64) = 128 J

Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.

Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)

Kinetic energy = Potential Energy

128 J = 39.2h

h = 3.26 m

h= 3.3 m (because of significant figures)

7 0
3 years ago
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