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kompoz [17]
3 years ago
8

Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.650 µC charge and flies due west at a speed of

540 m/s over the Earth's south magnetic pole, where the 8.00 ✕ 10−5 T magnetic field points straight up. What are the magnitude (in N) and direction of the magnetic force on the plane?
Physics
1 answer:
Ganezh [65]3 years ago
4 0

Answer : F=2808\times 10^{-11}\ N

Explanation :

It is given that,

Charge, q=0.650\ \mu C=0.65\times 10^{-6}\ C

Velocity of Aircraft, v=540\ m/s  (in west)

Magnetic field, B=8 \times 10^{-5}\ T ( in north )

Using the relation as :

F=q(v\times B)

Magnetic force is ,

F=0.65\times 10^{-6}\ C\times 540\ m/s\times 8 \times 10^{-5}\ T

F=2808\times 10^{-11}\ N

Using Right hand thumb rule, the direction of force is in the plane perpendicular to the velocity and the magnetic field i.e. -\hat{k}.

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Answer:

x₁ = 345100 km

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The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

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F₂  (force between mn and the object)

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F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

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Mt/ x₁²  =  Ml/ x₂²

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x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

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Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

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x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

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