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kkurt [141]
3 years ago
8

An electron moving with a velocity v⃗ = 5.0 × 107 m/s i^ enters a region of space where perpendicular electric and a magnetic fi

elds are present. The electric field is E⃗ = 104 V/m j^. What magnetic field will allow the electron to go through the region without being deflected?
Physics
1 answer:
stiv31 [10]3 years ago
3 0

Answer:

Magnetic field, B=2\times 10^{-4}\ T

Explanation:

Given that,

Velocity of electron, v=5\times 10^7\ m/s

It enters  a region of space where perpendicular electric and a magnetic fields are present.

Magnitude of electric field, E=10^4\ V/m

We need to find the magnetic field will allow the electron to go through the region without being deflected.

Magnetic force on the electron, F_m=qvB\ sin\theta.......(1)

Electric force on the electron, F = q E........(2)

From equation (1) and (2) we get:

qvB\ sin\theta=qE

B=\dfrac{E}{v}

B=\dfrac{10^4\ V/m}{5\times 10^7\ m/s}

B = 0.0002 T

or

B=2\times 10^{-4}\ T

Hence, this is the required solution.

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An object of volume 0.0882 m3 is
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The density of the fluid is 776.3 m^{-3}

<u>Explanation:</u>

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3 0
2 years ago
The Assignment: A fixed quantity of an ideal gas (R 0.28 kJ/kgK; Cv-0.71kJ/kgK) is expanded from an initial condition of 35 bar,
Nikolay [14]

Answer:

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q = 232.2 kJ/kg

Explanation:

The index of expansion is the relationship of pressures:

pi/pf

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p1*v1/T1 = p2*v2/T2

p2 = p1*v1*T2/(T2*v2)

500 C = 773 K

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p2 = 35*0.1*773/(293*1.3) = 7.1 bar

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The first law of thermodynamics

q = l + Δu

The work will be the expansion work

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35 bar = 3500000 Pa

7.1 bar = 710000 Pa

q = p2*v2 - p1*v1 + Δu

q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg

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