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FinnZ [79.3K]
2 years ago
15

We have a long wire with a circular cross section and radius a = 2.40 cm. The current density in this wire is uniform, with a to

tal current of 3.00 A. Find the magnitude of the magnetic field at a distance of 0.72 cm from the center axis. Treat the wire as a cylinder.
Physics
1 answer:
Alika [10]2 years ago
7 0

Answer:

The magnitude of the magnetic field is 7.49x10⁻⁶T

Explanation:

The magnetic field is:

B=(\frac{\mu i}{2\pi R^{2} } )r

Where

i = current = 3 A

R = radius = 2.4 cm = 0.024 m

r = distance = 0.72 cm = 7.2x10⁻³m

Replacing:

B=(\frac{4\pi x10^{-7} *3}{2\pi *0.024^{2} } )*7.2x10^{-3}= 7.49x10^{-6} T

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Answer:

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2 years ago
Two spherical shells have a common center. A -2.1 10-6 C charge is spread uniformly over the inner shell, which has a radius of
julsineya [31]

Answer:

a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing

b)  E_total = 1.89 10⁶ N / C, field is incoming radial

c) E_total = 0

Explanetion:

For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is

                Ф = E .dA = q_{int} /ε₀

inside the spherical shell there are no charges

The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.

We have two sphere shells with radii 0.050m and 0.15m respectively

a) point where you want to know the electric field d = 0.20 m

shell 1

the point is on the outside,d>ro,  therefore we can consider the charge to be concentrated in the center

            E₁ = k q₁ / d²

             

shell 2

the point is on the outside,d>ro

             E₂ = k q₂ / d²

the total camp is

              E_total = -E₁ + E₂

              E_total = k ( \frac{-q_1 + q_2}{d^2})

              E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²

              E_total = 6,525 10⁵ N /C

The field direction is radial and outgoing ti the shells

b) the calculation point is d = 0.10m

shell 1

point outside the shell d> ro

                 E₁ = k q₁ / d²

shell 2

the point is inside the shell d <ro

Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero

                E₂ = 0

               

                 E_total = E₁

                 E_total = k q₁ / d²

                 E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²

                 E_total = 1.89 10⁶ N / A

As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1

c) the point of interest d = 0.025 m

shell 1

point  is inside the shell d< ro

                 

as there are no charges inside

                     E₁ = 0

shell 2

point is inside the radius of the shell d <ro

                    E₂ = 0

the total field is

                    E_total = 0

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A 39.3 g glass thermometer reads 22.0oC before it
ratelena [41]

Answer:

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Explanation:

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Let the specific heat of water be 4.186 j/g °C

Let the water density be 1kg/L

136 mL of water = 0.136L of water = 0.136 kg of water = 136 g of water

Since the change of temperature on the glass thermometer is 43.6 - 22 = 21.6 C. We can then calculate the heat energy absorbed to it:

E = m_gc_g \Delta T = 39.3 * 0.84 * 21.6 = 713.06 J

Assume no energy is lost to outside, by the law of energy conservation, this heat energy would come from water

E = m_wc_w(T - T_w) = 713.06

136*4.186(T - 43.6) = 713.06

T - 43.6 = \frac{713.06}{136*4.186} = 1.25

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