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matrenka [14]
3 years ago
10

A proton moves through a region of space where there is a magnetic field B⃗ =(0.64i+0.40j)T and an electric field E⃗ =(3.3i−4.5j

)×103V/m. At a given instant, the proton's velocity is v⃗ =(6.6i+2.8j−4.8k)×103m/s.
Determine the components of the total force on the proton.

Express your answers using two significant figures. Enter your answers numerically separated by commas.
Physics
1 answer:
fenix001 [56]3 years ago
7 0

Answer:

F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k

Explanation:

When a charge is moving in constant magnetic field and electric field both then the net force on moving charge is vector sum of force due to magnetic field and electric field both

so first the force on the moving charge due to electric field is given by

\vec F_e = q\vec E

\vec F_e = (1.6 \times 10^{-19})(3.3 \hat i - 4.5 \hat j) \times 10^3

\vec F_e = (5.28 \times 10^{-16}) \hat i - (7.2 \times 10^{-16}) \hat j

Now force on moving charge due to magnetic field is given as

\vec F_b = q(\vec v \times \vec B)

\vec F_b = (1.6 \times 10^{-19})((6.6 \hat i+2.8 \hat j−4.8 \hat k) \times 10^3 \times (0.64 \hat i + 0.40 \hat j) )

\vec F_b = (4.22 \times 10^{-16})\hat k - (2.87 \times 10^{-16})\hat k - (4.92 \times 10^{-16})\hat j + (3.07 \times 10^{-16}) \hat i

\vec F_b = (3.07\times 10^{-16})\hat i - (4.92 \times 10^{-16})\hat j + (1.35 \times 10^{-16})\hat k

Now net force due to both

F = F_e + F_b

F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k

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The value of the force, F₀, at equilibrium is equal to the horizontal

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Response:

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<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

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Therefore;

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M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}

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F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2}  \approx  \mathbf{0.377  \cdot M \cdot g}

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