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svlad2 [7]
3 years ago
8

A cosmic ray electron moves at 7.65 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where field strength is

1.10 ✕ 10−5 T. What is the radius (in m) of the circular path the electron follows?
Physics
2 answers:
mariarad [96]3 years ago
8 0

Answer:

Radius = 3.96m

Explanation:

In a cyclotron motion, the radius of a charged particle path in a magnetic feild is given by:

r = mv/qB

Where r = radius

m = mass of particle= 9.1×10^-31kg

q = charged electron = 1.6×10^-19C

B = magnetic feid = 7.65×10^6T

V = velocity = 7.65×10^6

r = (9.1×10^-31)×(7.65×10^6) / (1.6×10^-19)(1.10 ×10^-5)

r = (6.9615×10^-24)/(1.76 ×10^-24)

r = 3.96m

diamong [38]3 years ago
3 0

Answer:

Explanation:

Force of a magnetic field, Fm = q × v × B

Centipetal force, Fa = (M × v^2)/r

Therefore, Fm = Fa

q × v × B = (M × v^2)/r

r = (m × v)/(q × B)

Where,

r = radius

m = mass of electron

= 9.1 × 10^-31 kg

q = electron charge

= 1.602 × 10^-19 C

B = magnetic field

= 1.10 ✕ 10^−5 T

v = velocity

= 7.65×10^6 m/s

r = (9.1 × 10^-31) × (7.65 × 10^6) / (1.6 × 10^-19) × (1.10 × 10^-5)

= 3.95 m

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Which of the following statements about electromagnetic waves in free space are true? (There could be more than one correct choi
Vanyuwa [196]

Answer:

The electric field carries the same mount of energy as the magnetic field

The frequency of the magnetic field is the same as the frequency of the electric field.

Explanation:

Let's analyze each option in detail:

The electric and magnetic fields have equal amplitudes. --> FALSE. In fact, the amplitudes of the electric and magnetic field are related by the equation:

E=cB

where

E is the amplitude of the electric field

B is the amplitude of the magnetic field

c is the speed of light

Therefore, from the equation we see that the amplitude of the electric field is much larger than that of the magnetic field.

The electric field carries the same mount of energy as the magnetic field. --> TRUE.

The energy carried by the electric field is:

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where \epsilon_0 is the vacuum permittivity.

The energy carried by the magnetic field is:

u_B = \frac{1}{2\mu_0}B^2

where \mu_0 is the vacuum permeability.

Given the following relationship:

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We can write

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u_E = \frac{1}{2}\epsilon_0 (\frac{1}{\sqrt{\epsilon_0 \mu_0}}B)^2=\frac{1}{2\mu_0}B^2

which means u_E=u_B.

The electric field carries more energy than the magnetic field. --> FALSE, because in disagreement with the calculations above.

The frequency of the magnetic field is the same as the frequency of the electric field. --> TRUE. In an electromagnetic waves, electric field and magnetic field oscillate at the same frequency.

The frequency of the electric field is higher than the frequency of the magnetic field. --> FALSE, because in disagreement with the previous statement.

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Bezzdna [24]

Answer:

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