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2 years ago

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An ion with charge of Q = +3.2 x 10-19 C is in a region where a uniform electric field of magnitude E = 5.0 X 105 V/m is perpend

icular to a uniform magnetic field B = 0.80 T. If the ion's acceleration is zero then what is its speed v? a) v = 1.6 x 10-5 m/s b) v = 4.0 x 105 m/s c) v = 6.3 x 105 m/s d) v = 0 m/s

1 answer:

san4es73 [151]2 years ago

8 0

Answer:

option C

Explanation:

given,

Q = +3.2 x 10⁻¹⁹ C

E = 5.0 X 10⁵ V/m

B = 0.80 T

ion's acceleration is zero

when acceleration is zero the magnitude of both the forces becomes equal.

q E = q V B

v = $\dfrac{E}{B}$

v= $\dfrac{5 \times 10^5}{0.80}$

v = 6.25 × 10⁵ m/s ≈ 6.3 × 10⁵ m/s

hence, the correct answer is option C

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It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

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But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

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That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

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<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

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For the case of the particle in the D Ring, the same approach used above can be followed

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