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9966 [12]
3 years ago
8

A cosmic-ray proton in interstellar space has an energy of 59 MeV and executes a circular orbit with a radius equal to that of M

ercury’s orbit (5.8 × 1010 m) around the Sun. What is the magnetic field in that region of space? The proton has a charge of 1.60218 × 10−19 C and a mass of 1.67262 × 10−27 kg. Answer in units of T
Physics
1 answer:
makkiz [27]3 years ago
8 0

Answer:

Explanation:

Let mass and velocity of proton be m and v .

1/2 m v² = 59 x 10⁶ e V

= 59 x 10⁶ x 1.6 x 10⁻¹⁹ J

= 94.4 x 10⁻¹³ J

mv² = 188.8 x 10⁻¹³ J

v² =  188.8 x 10⁻¹³ / m

= 188.8 x 10⁻¹³ / 1.67262 x 10⁻²⁷

= 112.8768 x 10¹⁴

v = 10.62 x 10⁷ m / s

In circular path of proton , magnetic force equals centripetal force .

m v² / r = B q v , B is magnetic field , q is charge on proton , r is radius of circular path .

188.8 x 10⁻¹³ / 5.8 x 10¹⁰ = B x 1.6 x 10⁻¹⁹ x  10.62 x 10⁷

B = 1.9157 x 10⁻¹¹ T.

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Answer:

the electric field at point A is

E = 5.5 ×10¹³N/C(-x direction)

Explanation:

given

electrostatics constant k = 9.0×10⁹

charge q = 31.7mC= 31.7×10⁻³C

distance r = 2.80mm

distance from midpoint to point A = 2.00mm

attached is the diagram of the solution, describing the position of the charge

note x = r/2, where x is the distance from midpoint of r to the particle

using Pythagoras theorem as in the attachment, x = 2.44mm= 2.44×10⁻³m

the electric field at point A is given as

vector <em>E </em>= 2E×cos θ( -x direction)

recall E =kq/x²

where k is the electrostatics constant = 1/4πε₀

where ε₀ is permittivity of free space

therefore using E =2{kq/x²}cosθ

∴cosθ = adjacent/hypotenuse

cosθ=1.40/2.44

E =2 {(9.0×10⁹ × 31.7×10⁻³) ÷ (2.44×10⁻³)²}×(1.40/2.44)(-x)

E=2{4.79×10¹³}×(0.574)(-x)

E = 2×2.75 ×10¹³N/C(-x direction)

Vector <em>E= </em>5.5 ×10¹³N/C(-x direction)

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Answer:

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A rope is tied to a box and used to pull the box 2.3 m along a horizontal floor. The rope makes an angle of 30∘ with the horizon
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Answer:

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