Physics: The voltage in an EBW operation is 45 kV. The beam current is 50 milliamp. The electron beam is focused on a circular area that is 0.50 mm in diameter. The heat transfer factor is 0.87. Calculate the average power density in the area in watt/mm2.

The voltage in an EBW operation is 45 kV. The beam

zemathes

21.08.2021

P_d=6203.223062W/mm^2

Explanation:

From the question we are told that:

Voltage V=45kV

Current I=50mAmp

Diameter d=0.50mm

Heat transfer factor $\mu= 0.87.$

Generally the equation for Power developed is mathematically given by

$P=VI\\\\P=45*10^3*50*10^{-3}$

P=2.250

Therefore

Power in area

P_a=1400*0.87

P_a=1218watt

Power Density

$P_d=\frac{P_a}{Area}$

$P_d=\frac{1218}{\pi(0.5^2/4)}$

P_d=6203.223062W/mm^2

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quinnie48

02.08.2022

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

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