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Sagot :
Answer: The mutual inductance of these solenoids is [tex]2.88 \times 10^{-7} H[/tex].
Explanation:
Given: Length = 50.0 cm (1 cm = 0.01 m) = 0.50 m
[tex]N_{1}[/tex] = 6750
[tex]N_{2}[/tex] = 15
Radius = [tex]\frac{0.120 cm}{2} = 0.6 cm = 6 \times 10^{-4} m[/tex]
As inner of a solenoid resembles the shape of a circle. So, its area is calculated as follows.
[tex]Area = \pi \times r^{2} = \pi \times (6 \times 10^{-4})^{2}[/tex]
Formula used to calculate mutual conductance of two solenoids is as follows.
[tex]M = \frac{\mu_{o} \times A \times N_{1} \times N_{2}}{l}[/tex]
where,
M = mutual conductance
A = area
[tex]\mu_{o}[/tex] = relative permeability = [tex]4 \pi \times 10^{-7} Tm/A[/tex]
[tex]N_{1}[/tex] = no. of coils in outer solenoid
[tex]N_{2}[/tex] = no. of coils in inner solenoid
l = length
Substitute the values into above formula as follows.
[tex]M = \frac{\mu_{o} \times A \times N_{1} \times N_{2}}{l}\\= \frac{4 \pi \times 10^{-7} Tm/A \times \pi (6 \times 10^{-4})^{2} \times 6750 \times 15}{0.5 m}\\= 2.88 \times 10^{-7} H[/tex]
Thus, we can conclude that the mutual inductance of these solenoids is [tex]2.88 \times 10^{-7} H[/tex].
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