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Chapter 26: Problem 69

Derive Equation 26.21 for the solenoid field by considering the solenoid to be made of infinitesimal current loops. Use Equation 26.9 for the loop fields, and integrate over all loops.

Short Answer

Expert verified
The magnetic field of a long solenoid, as given by Equation 26.21, is derived to be \( B = \frac{{\mu_0 \cdot N \cdot I}}{2} \).

Step by step solution

01

State Equation 26.9

We start with Equation 26.9, which describes the magnetic field, \(B\), at the center of a small current loop with radius \(r\) and current \(I\). It is given by: \[ B = \frac{{\mu_0 \cdot I \cdot R^2}}{{2(R^2 + X^2)^(3/2)}} \] where \(R\) is the radius of the loop, \(X\) is the distance from the center of the loop to the point where you want to calculate \(B\), and \(\mu_0\) is the permeability of free space.
02

Set up the integral for the solenoid

We imagine the solenoid as being composed of many infinitesimal current loops stacked long it's length. Since we're considering the field inside the solenoid, \(X = 0\). The expression simplifies to: \[ B = \frac{{\mu_0 \cdot I \cdot R^2}}{{2R^3}} = \frac{{\mu_0 \cdot I}}{{2R}}\] For the whole solenoid, we sum up the contribution of each loop by integrating over the length of the solenoid, \(L\). The magnetic field, \(B\), is then given by: \[ B = \int \frac{{\mu_0 \cdot I}}{{2R}} ds \] where \(ds\) denotes the thickness of each infinitesimal loop.
03

Perform the integration

Performing the integration, and considering \(N\) turns of the coil in the solenoid with length \(L\), means replacing \(\frac{I}{R}\) by \(N\cdot I/L\). This gives us \[ B = \int \frac{{\mu_0 \cdot N \cdot I}}{{2L}} ds \]. Considering the coil is uniformly wound and integrated over the total length \(L\), we get, \[ B = \frac{{\mu_0 \cdot N \cdot I}}{{2L}} \cdot L \]
04

Simplify the result

After simplifying, we get the desired Equation 26.21: \[ B = \frac{{\mu_0 \cdot N \cdot I}}{2} \] which describes the magnetic field inside a long solenoid. Note that this equation implies that the magnetic field is uniform inside the solenoid and is directed along the axis of the solenoid.

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