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3 years ago

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This problem has been solved! See the answer A proton with a speed of 3.5x10^6 m/s is shot into a region between two plates that

are separated by distance of 0.23 m. A magnetic field exists between the plates, and it is perpendicular to the velocity of the proton. What must be the magnitude of the magnetic field so the proton just misses colliding with the opposite plate?

1 answer:

DIA [1.3K]3 years ago

7 0

Answer:

The magnitude of the magnetic field 'B' is 0.16 Tesla.

Explanation:

The magnitude of the magnetic field 'B' can be determined by;

B = $\frac{mV}{qR}$

where: m is the mass of proton, V is its speed , q is the charge of proton and R is the distance between the plates.

Given that: speed 'V' of the proton = 3.5 × $10^{6} ms^{-2}$ , distance 'R' between the plates = 0.23m, the charge 'q' on proton = 1.9 × $10^{-19}$ C and mass of proton = 1.67 × $10^{-27}$ Kg.

Thus,

B = (1.67 × $10^{-27}$ × 3.5 × $10^{6}$ ) ÷ (1.6 × $10^{-19}$ × 0.23)

= $\frac{5.845 * 10^{-21} }{3.68 * 10^{-20} }$

= 0.15883

B = 0.16 Tesla

The magnitude of the magnetic field 'B' is 0.16 Tesla.

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The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

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By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

$P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2}$ (1)

Where:

$P_{1}$ , $P_{2}$ - Pressures of the water on the first and second floors, measured in pascals.

$\rho$ - Density of water, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Speed of the water on the first and second floors, measured in meters per second.

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Now we clear the final speed of the water flow:

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$v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}$

$v_{2} \approx 13.100\,\frac{m}{s}$

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