Respuesta :
Answer:
∴Inductance of solenoid A is [tex]\frac18[/tex] of inductance of solenoid B.
Explanation:
Inductance of a solenoid is
[tex]L=N\frac\phi I[/tex]
[tex]=N\frac{B.A}{I}[/tex]
[tex]=N\frac{\mu_0NI}{l.I}A[/tex]
[tex]=\frac{\mu_0N^2A}{l}[/tex]
[tex]=\frac{\mu_0N^2}{l}.\pi(\frac d2)^2[/tex]
[tex]=\mu_0\pi\frac{N^2d^2}{4l}[/tex]
N= number of turns
[tex]l[/tex] = length of the solenoid
d= diameter of the solenoid
A=cross section area
B=magnetic induction
[tex]\phi[/tex] = magnetic flux
[tex]I[/tex]= Current
Given that, Solenoid A has total number of turns N, length L and diameter D
The inductance of solenoid A is
[tex]=\mu_0\pi\frac{N^2D^2}{4L}[/tex]
Solenoid B has total number of turns 2N, length 2L and diameter 2D
The inductance of solenoid B is
[tex]=\mu_0\pi\frac{(2N)^2(2D)^2}{4.2L}[/tex]
[tex]=\mu_0\pi\frac{16 N^2D^2}{4.2L}[/tex]
Therefore,
[tex]\frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=\frac{\mu_0\pi\frac{N^2D^2}{4L}}{\mu_0\pi\frac{16 N^2D^2}{4.2L}}[/tex]
[tex]\Rightarrow \frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=\frac18[/tex]
[tex]\Rightarrow {\textrm{Inductance of A}}=\frac18\times {\textrm{Inductance of B}}[/tex]
∴Inductance of solenoid A is [tex]\frac18[/tex] of inductance of solenoid B.
Hi there!
We can begin by calculating the inductance of a solenoid.
Recall:
[tex]L = \frac{\Phi _B}{i}[/tex]
L = Inductance (H)
φ = Magnetic Flux (Wb)
i = Current (A)
We can solve for the inductance of a solenoid. We know that its magnetic field is equivalent to:
[tex]B = \mu _0 \frac{N}{L}i[/tex]
And that the magnetic flux is equivalent to:
[tex]\Phi _B = \int B \cdot dA = B \cdot A[/tex]
Thus, the magnetic flux is equivalent to:
[tex]\Phi _B = \mu _0 \frac{N}{L}iA[/tex]
The area for the solenoid is the # of loops multiplied by the cross-section area, so:
[tex]A_{total}= N * A[/tex]
[tex]\Phi _B = \mu _0 \frac{N^2}{L}iA[/tex]
Using this equation, we can find how it would change if the given parameters are altered:
[tex]\Phi_B ' = \mu_0 \frac{(2N)^2}{2L} i * 4A[/tex]
**The area will quadruple since a circle's area is 2-D, and you are doubling its diameter.
[tex]\Phi'_B = \frac{4}{2} * 4(\mu_0 \frac{N}{L}iA) = 8\mu_0 \frac{N}{L}iA[/tex]
Thus, Solenoid B is 8 times as large as Solenoid A.
Solenoid A is 1/8 of the inductance of solenoid B.
