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lisov135 [29]
3 years ago
9

A square loop of wire lies in the x, y plane with opposite corners at (0, 0) and (a, a). The loop carries a current I directed t

owards the origin along the y-axis and away from the origin along the x-axis. Calculate the cartesian components of the magnetic field B generated by the loop at the point (x, y, z) = (0, 0, a), in terms of a, I, and fundamental constants.

Physics
1 answer:
Ket [755]3 years ago
5 0

Answer:

Please refer to the figure.

Explanation:

The magnitude of the magnetic field can be found by Biot-Savart Law. We should divide the loop into four components. Each component has a similar solution but their directions are quite different.

The directions can be found by right-hand rule. Point your index finger into the direction of current, point your middle finger towards the target point (0,0,a). Your thumb will show you the direction of magnetic field.

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The vapor pressure of benzene, C6H6, is 40.1 mmHg at 7.6°C. What is its vapor pressure at 60.6°C? The molar heat of vaporization
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The vapor pressure at 60.6°C is 330.89 mmHg

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Applying Clausius Clapeyron Equation

ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]

Where;

P₂ is the final vapor pressure of benzene = ?

P₁ is the initial vapor pressure of benzene = 40.1 mmHg

T₂ is the final temperature of benzene = 60.6°C = 333.6 K

T₁ is the initial temperature of benzene = 7.6°C = 280.6 K

ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol

R is gas rate = 8.314 J/mol.k

ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65  (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg

Therefore, the vapor pressure at 60.6°C is 330.89 mmHg

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