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pashok25 [27]
3 years ago
15

An electron is moving directly toward you in a horizontal path when it suddenly enters a uniform magnetic field that is either v

ertical or horizontal.
If the electron begins to curve upward in its motion just after it enters the field, you can conclude that the direction of the magnetic field is
If the electron begins to curve upward in its motion just after it enters the field, you can conclude that the direction of the magnetic field is
upward.
downward.
to your left.
to your right.
Physics
1 answer:
blagie [28]3 years ago
3 0

Answer:

To your left

Explanation:

The direction of the force exerted on charged particle due to a magnetic field is given by the right-hand-rule, where:

- The index finger indicates the direction of motion of the electron

- the middle finger gives the direction of the magnetic field

- the thumb gives the direction of the force if the particle is positively charged - otherwise, the direction is reversed

in this case, we have an electron (so, a negatively charged particle):

- The direction of motion (index finger) is horizontal, toward you

- The electron begins to curve upward as it enters the field, so this means that the force exerted on the electrons is upward --> the thumb must point downward (because the electron is negatively charged)

- The index finger gives us the direction of the magnetic field: therefore, to your left.

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When can get relive from stress by talking about our thoughts or worries with an individual. The other person might help us solve our problems or he might suggest us in a better way.

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A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.
cupoosta [38]

Answer:

a) The x and y components of the momentum are 8.731\,\frac{kg\cdot m}{s} and -11.661\,\frac{kg\cdot m}{s}, respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

\vec p = m\cdot \vec v (1)

Where:

\vec p - Vector momentum, measured in kilogram-meters per second.

m - Mass of the particle, measured in kilograms.

\vec v - Vector velocity, measured in meters per second.

If we know that m = 2.93\,kg and \vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right], then the momentum is:

\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]

\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]

The x and y components of the momentum are 8.731\,\frac{kg\cdot m}{s} and -11.661\,\frac{kg\cdot m}{s}, respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}} (2)

\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right) (3)

Where:

\|\vec p\| - Magnitude of momentum, measured in kilogram-meters per second.

\theta - Direction of momentum, measured in sexagesimal degrees.

If we know that p_{x} = 8.731\,\frac{kg\cdot m}{s} and p_{y} = -11.661\,\frac{kg\cdot m}{s}, then the magnitude and direction of momentum are, respectively:

\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}

\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}

\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)

\theta \approx 306.823^{\circ}

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

6 0
3 years ago
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