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Svetlanka [38]
3 years ago
11

At one instant, an electron (charge = –1.6 x 10–19 C) is moving in the xy plane, the components of its velocity being vx = 5.0 x

105 m/s and vy = 3.0 x 105 m/s. A magnetic field of 0.80 T is in the positive y direction. At that instant, what is the magnitude of the magnetic force on the electron?
Physics
1 answer:
Tomtit [17]3 years ago
3 0

Answer:

The direction of force is along negative Z axis and the magnitude of force is

6.4 x 10^-14 N.

Explanation:

q = - 1.6 x 10^-19 c

vx = 5 x 10^5 m/s, vy = 3 x 10^5 m/s, B = 0.8 T along Y axis

The velocity vector is given by

v = 5 x 10^5 i + 3 x 10^5 j

B = 0.8 j

Force on a charged particle place in a magnetic field is given by

F = q (v x B)

F = -1.6 x 10^-19 {(5 x 10^5 i + 3 x 10^5 j) x (0.8 j)}

F = - 1.6 x 10^-19 (5 x 0.8 x 10^5 k)

F = - 6.4 x 10^-14 k

The direction of force is along negative Z axis and the magnitude of force is

6.4 x 10^-14 N.

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The anserrs to the question are

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(b) The pressure of the resulting air stream is 14 bar

Explanation:

Here the two streams of gases meet ad form a single stream

The steady-flow energy equation can be implemented at the mixing point of the to streams as follows

mₐhₐ₁ + mₙ hₙ₁ + Q° + W° = mₐhₐ₂ + mₙhₙ₂

Where the flow is adiabatic, we have

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mₐhₐ₁ + mₙ hₙ₁ = mₐhₐ₂ + mₙhₙ₂ where h = cp×T we have

mₐcpₐ×Tₐ + mₙcpₙ×Tₙ  = mₐcpₐ×T + mₐcpₙ×T

Therefore the final temperature T is given by

T = \frac{m_ac_{pa}T_a + m_nc_{pn}T_n}{m_ac_{pa} + m_nc_{pn}}  for the same kind of gas cpₐ =cpₙ therefore

T = \frac{m_aT_a + m_nT_n}{m_a + m_n} where mₐ and mₙ are mass flow rate

Therefore we have where Tₐ = = 900 K and Tₙ = 400 K and mₙ = 2mₐ gives

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