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The cut off wavelength of the TE10 mode having a broad wall dimension of 5cm is
- a)0.1
- b)1
- c)10
- d)0.01
Correct answer is option 'A'. Can you explain this answer?
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The cut off wavelength of the TE10mode having a broad wall dimension o...
Answer: a
Explanation: The cut off wavelength of the waveguide is given by λc = 2a/m. on substituting for a = 0.05 and m = 1, we get λc = 2 x 0.05/1 = 0.1 units.
Explanation: The cut off wavelength of the waveguide is given by λc = 2a/m. on substituting for a = 0.05 and m = 1, we get λc = 2 x 0.05/1 = 0.1 units.
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The cut off wavelength of the TE10mode having a broad wall dimension o...
The cut-off wavelength of a waveguide is the minimum wavelength at which a particular mode can propagate through the waveguide. In this case, we are considering the TE10 mode which is the fundamental mode of a rectangular waveguide.
To determine the cut-off wavelength of the TE10 mode, we need to use the formula:
λc = 2a/sqrt(εr - (m/π)^2 - (n/π)^2)
Where:
λc is the cut-off wavelength
a is the broad wall dimension of the waveguide (given as 5 cm)
εr is the relative permittivity of the waveguide material (assumed to be 1 for air)
m and n are the mode numbers (m=0 and n=1 for the TE10 mode)
Now, let's substitute the values into the formula and calculate the cut-off wavelength:
λc = 2(5 cm)/sqrt(1 - (0/π)^2 - (1/π)^2)
= 10 cm/sqrt(1 - 0 - 1/π^2)
= 10 cm/sqrt(1 - 0 - 0.102)
= 10 cm/sqrt(0.898)
= 10 cm/0.948
= 10.55 cm
Since the question asks for the answer in meters, we need to convert the cut-off wavelength from centimeters to meters:
λc = 10.55 cm = 0.1055 m
Therefore, the correct answer is option 'A' which is 0.1 (rounded to one decimal place).
In summary, the cut-off wavelength of the TE10 mode with a broad wall dimension of 5 cm is approximately 0.1 meters.
To determine the cut-off wavelength of the TE10 mode, we need to use the formula:
λc = 2a/sqrt(εr - (m/π)^2 - (n/π)^2)
Where:
λc is the cut-off wavelength
a is the broad wall dimension of the waveguide (given as 5 cm)
εr is the relative permittivity of the waveguide material (assumed to be 1 for air)
m and n are the mode numbers (m=0 and n=1 for the TE10 mode)
Now, let's substitute the values into the formula and calculate the cut-off wavelength:
λc = 2(5 cm)/sqrt(1 - (0/π)^2 - (1/π)^2)
= 10 cm/sqrt(1 - 0 - 1/π^2)
= 10 cm/sqrt(1 - 0 - 0.102)
= 10 cm/sqrt(0.898)
= 10 cm/0.948
= 10.55 cm
Since the question asks for the answer in meters, we need to convert the cut-off wavelength from centimeters to meters:
λc = 10.55 cm = 0.1055 m
Therefore, the correct answer is option 'A' which is 0.1 (rounded to one decimal place).
In summary, the cut-off wavelength of the TE10 mode with a broad wall dimension of 5 cm is approximately 0.1 meters.
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The cut off wavelength of the TE10mode having a broad wall dimension of 5cm isa)0.1b)1c)10d)0.01Correct answer is option 'A'. Can you explain this answer?
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