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3 years ago

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A 40 cm wire with a radius of 3 cm is oriented along the y axis and carries a current of 2 A. What is the magnitude of the magne

tic field 2 cm from the center of the wire? A 40 cm wire with a radius of 3 cm is oriented along the y axis and carries a current of 2 A. What is the magnitude of the magnetic field 4 cm from the center of the wire

1 answer:

ZanzabumX [31]3 years ago

7 0

Answer:

a) B = 1.99 x 10⁻⁴ Tesla

b) B = 0.88 x 10⁻⁴ Tesla

Explanation:

According to Biot - Savart Law, the magnetic field due to a currnt carrying straight wire is given as:

B = μ₀ I L/4πr²

where,

μ₀ = permebility of free space = 1.25 x 10⁻⁶ H m⁻¹

I = current = 2 A

L = Length of wire = 40 cm = 0.4 m

a)

r = radius of magnetic field = 2 cm = 0.02 m

Therefore,

B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.02 m)²

<u>B = 1.99 x 10⁻⁴ Tesla</u>

<u></u>

b)

r = radius of magnetic field = 3 cm = 0.03 m

Therefore,

B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.03 m)²

<u>B = 0.88 x 10⁻⁴ Tesla</u>

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A domestic water heater holds 189 L of water at 608C, 1 atm. Determine the exergy of the hot water, in kJ. To what elevation, in

Gekata [30.6K]

A.

The energy of the hot water is 482630400 J

Using Q = mcΔT where Q = energy of hot water, m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 189 L = 0.189 m³,

c = specific heat capacity of water = 4200 J/kg-°C and ΔT = temperature change of water = T₂ - T₁ where T₂ = final temperature of water = 608 °C. If we assume the water was initially at 0°C, T₁ = 0 °C. So, the temperature change ΔT = 608 °C - 0 °C = 608 °C

Substituting the values of the variables into the equation, we have

Q = mcΔT

Q = ρVcΔT

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Q = 482630400 J

So, the energy of the hot water is 482630400 J

B.

The elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.

Using the equation for gravitational potential energy ΔU = mgΔh where m = mass of object = 1000 kg, g = acceleration due to gravity = 9.8 m/s² and Δh = h - h' where h = required elevation and h' = zero level elevation = 0 m

Since the energy of the mass equal the energy of the hot water, ΔU = 482630400 J

So, ΔU = mgΔh

ΔU = mg(h - h')

making h subject of the formula, we have

h = h' + ΔU/mg

Substituting the values of the variables into the equation, we have

h = h' + ΔU/mg

h = 0 m + 482630400 J/(1000 kg × 9.8 m/s²)

h = 0 m + 482630400 J/(9800 kgm/s²)

h = 0 m + 49248 m

h = 49248 m

So, the elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.

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Given:

Inductance, L = 150 mH

Capacitance, C = 5.00 mF

$\Delta V_{max}$ = 240 V

frequency, f = 50Hz

$I_{max}$ = 100 mA

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To calculate the parameters of the given circuit series RLC circuit:

angular frequency, $\omega$ = $2\pi f = 2\pi \times50 = 100\pi$

a). Inductive reactance, $X_{L}$ is given by:

$\X_{L} = \omega L = 100\pi \times 150\times 10^{-3} = 47.12\Omega$

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$Z = 2400\Omega$

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$R = 2399.5\Omega$

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$tan\phi = \frac{X_{L} - X_{C}}{R}$

$\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})$

$\phi =tan^{-1}( \frac{47.12 - 0.636}{2399.5}) = tan^{-1}{0.0.01937}$

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