Respuesta :
Answer:
The winding density of the solenoid, n = 104 turns/m
Explanation:
Given that,
Length of the solenoid, l = 0.7 m
Radius of the circular cross section, r = 5 cm = 0.05 m
Energy stored in the solenoid, [tex]E=6\ \mu J=6\times 10^{-6}\ J[/tex]
Current, I = 0.4 A
To find,
The winding density of the solenoid.
Solution,
The expression for the energy stored in the solenoid is given by :
[tex]U=\dfrac{1}{2}LI^2[/tex]
Where
L is the self inductance of the solenoid
[tex]L=\mu_on^2lA[/tex]
n is the winding density of the solenoid
[tex]n=\sqrt{\dfrac{2U}{\mu_oI^2l\pi r^2}}[/tex]
[tex]n=\sqrt{\dfrac{2\times 6\times 10^{-6}}{4\pi \times 10^{-7}\times 0.7\times (0.4)^2\pi (0.05)^2}}[/tex]
n = 104 turns/m
So, the winding density of the solenoid is 104 turns/m
Answer
The winding density of the solenoid is 104 turns/m.
(D) is correct option.
Explanation:
Given that,
Length = 0.700 m
Radius = 5.00 cm
Energy = 6.00 μJ
Current = 0.400 A
We need to calculate the density of the solenoid
Using formula of the energy store
[tex]E=\dfrac{B^2}{2\mu_{0}}Al[/tex]
Put the value of magnetic field
[tex]E=\dfrac{(\mu_{0}ni)^2}{2\mu_{0}}Al[/tex]
[tex]n=\sqrt(\dfrac{2E}{\mu_{0}i^2Al})[/tex]
Put the value into the formula
[tex]n=\sqrt{\dfrac{2\times6.00\times10^{-6}}{4\pi\times10^{-7}\times(0.4)^2\times\pi\times(5.00\times10^{-2})^2\times0.700}}[/tex]
[tex]n=104\ turns/m[/tex]
Hence, The winding density of the solenoid is 104 turns/m.
