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3 years ago

5

The two conducting rails in the drawing are tilted upwards so they each make an angle of 30.0° with respect to the ground. The v

ertical magnetic field has a magnitude of 0.045 T. The 0.19 kg aluminum rod (length = 1.6 m) slides without friction down the rails at a constant velocity. How much current flows through the rod?

1 answer:

dolphi86 [110]3 years ago

8 0

Answer:

The current flows through the rod is 14.9 A.

Explanation:

Given that,

Magnetic field = 0.045 T

Mass of aluminum rod = 0.19 kg

Length = 1.6 m

Angle = 30.0°

We need to calculate the force

Using resolving force

$F\cos\theta=mg\sin\theta$

$F=mg\tan\theta$

Put the value into the formula

$F=0.19\times9.8\times\tan30$

$F=1.075\ N$

We need to calculate the current flows through the rod

Using formula of magnetic force

$F=iLB$

$i=\dfrac{F}{LB}$

Put the value into the formula

$i=\dfrac{1.075}{1.6\times0.045}$

$i=14.9\ A$

Hence, The current flows through the rod is 14.9 A.

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Where:

$P$ - Pressure of the bubble, measured in kilopascals.

$V$ - Volume of the bubble, measured in cubic meters.

$n$ - Molar amount of the bubble, measured in kilomoles.

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Then, we eliminate the molar amount and the ideal gas constant by constructing the following relationship:

$\frac{P_{A}\cdot V_{A}}{T_{A}} = \frac{P_{B}\cdot V_{B}}{T_{B}}$ (Eq. 2)

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$P_{A}$ , $P_{B}$ - Pressure of the bubble at bottom and surface, measured in kilopascals.

$V_{A}$ , $V_{B}$ - Volume of the bubble at bottom and surface, measured in cubic meters.

$T_{A}$ , $T_{B}$ - Temperature of the bubble at bottom and surface, measured in Kelvin.

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$P_{A} = 101.325\,kPa+\left(1027\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{kg}{m^{3}} \right)\cdot (25\,m)\cdot \left(\frac{1}{1000}\,\frac{kPa}{Pa} \right)$

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$V_{B} = 4.372\,cm^{3}$

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$V_{B} = \frac{4\pi\cdot R^{3}}{3}$ (Eq. 3)

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