A long solenoid has a radius of \(3 \mathrm{~cm}, 5000\) turns \(/ \mathrm{m}\), and carries current \(I=0.25 \mathrm{~A} .\) The region \(0<\rho

Short Answer

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Question: Calculate the radius 'a' in a solenoid, given two different conditions on the total magnetic flux: (1) the total flux is 10μWb, (2) the flux is equally divided between two regions with different magnetic permeability. Answer: In both cases, the value of the radius 'a' is different. 1. For the total flux of 10μWb, the radius 'a' is approximately 1.4 x 10^-2 m. 2. For the equally divided magnetic flux between the two regions, the radius 'a' is approximately 2.2 x 10^-2 m.

Step by step solution

01

Write down the expression for the magnetic field in a solenoid

Since the magnetic field in the solenoid is parallel to the solenoid axis, we can consider only the axial component \(B_z\). The radial component \(B_\rho\) is irrelevant for the calculation of the magnetic flux. Using Ampere's Law, we find the axial magnetic field inside the solenoid (Region 1) given by: \[B_{z1} = \mu_{r1} \mu_0 n I\] Similarly, for Region 2, we have: \[B_{z2} = \mu_{r2} \mu_0 n I\]
02

Write down the expression for the magnetic flux in each region

The magnetic flux \(\Phi\) is given by the product of the magnetic field and the area through which it passes. Let's consider a cylindrical surface of radius \(\rho\) and height \(h\). The flux in each region can be defined as: \[\Phi_1 = B_{z1} (\pi(\rho^2-a^2)h)\] \[\Phi_2 = B_{z2} (\pi(a^2)h)\]
03

Solve the problem for case(a)

Given that the total flux is \(10\mu\text{Wb}\), we have: \[\Phi_{total} = \Phi_1 + \Phi_2 = 10\times10^{-6}\text{Wb}\] Substitute the expressions for \(B_{z1}\) and \(B_{z2}\) that we got in Step 1: \[(\mu_{r1}\mu_0nI)(\pi(\rho^2-a^2)h)+(\mu_{r2}\mu_0nI)(\pi(a^2)h)=10\times 10^{-6}\text{Wb}\] Using the given values, \(\mu_{r1}=5\), \(\mu_{r2}=1\), \(\rho=3\times10^{-2}\text{m}\), \(n=5000\text{turns/m}\), and \(I=0.25\text{A}\), we can solve for the unknown variable (a): \[a^2(4\pi\times10^{-7}\times 5000\times0.25)=5\times10^{-6}\] \[a^2 = \frac{5\times10^{-6}}{(4\pi\times10^{-7}\times 5000\times0.25)}\] Solving for "a", we get: \[a \approx 1.4\times10^{-2} \text{m}\]
04

Solve the problem for case(b)

In this case, it's given that the flux is equally divided between the regions \(0 < \rho < a\) and \(a < \rho < 3\text{cm}\). Therefore, we have: \[\Phi_1 = \Phi_2\] \[(\mu_{r1}\mu_0nI)(\pi(\rho^2-a^2)h) = (\mu_{r2}\mu_0nI)(\pi(a^2)h)\] Using the same given values as in case(a), we get: \[5(\rho^2-a^2) = a^2\] Since \(\rho = 3\times10^{-2}\text{m}\), we can solve for "a": \[a^2 = \frac{5(\rho^2)}{6}\] \[a^2 = \frac{5(3\times10^{-2})^2}{6}\] Solving for "a", we get: \[a \approx 2.2\times10^{-2} \text{m}\] The value of "a" found under both cases are as follows: Case(a): \(a \approx 1.4\times10^{-2} \text{m}\) Case(b): \(a \approx 2.2\times10^{-2} \text{m}\)

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