To find the magnetic moment of the solenoid based on the given magnetic flux, we can follow these steps:
### Step 1: Understand the relationship between magnetic flux, magnetic field, and area.
The magnetic flux (Φ) through a surface is given by the formula:
\[
\Phi = B \cdot A \cdot \cos(\theta)
\]
Where:
- \(B\) is the magnetic field,
- \(A\) is the area,
- \(\theta\) is the angle between the magnetic field and the normal to the surface.
Since the magnetic field and the area vector are parallel in this case, \(\theta = 0\) degrees, and thus \(\cos(0) = 1\).
So, we can simplify the equation to:
\[
\Phi = B \cdot A
\]
### Step 2: Determine the magnetic field inside the solenoid.
The magnetic field inside a long solenoid is given by:
\[
B = \mu_0 \cdot n \cdot I
\]
Where:
- \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\) (permeability of free space),
- \(n\) is the number of turns per unit length,
- \(I\) is the current flowing through the solenoid.
### Step 3: Relate magnetic moment to magnetic flux.
The magnetic moment (\(m\)) of the solenoid can be expressed as:
\[
m = n \cdot I \cdot A
\]
Where \(A\) is the cross-sectional area of the solenoid.
### Step 4: Set up the relationship between magnetic flux and magnetic moment.
From the equations for flux and magnetic moment, we can relate them:
\[
\Phi = B \cdot A = \mu_0 \cdot n \cdot I \cdot A
\]
Dividing both sides by \(A\):
\[
\frac{\Phi}{A} = \mu_0 \cdot n \cdot I
\]
Now, we can express \(n\) in terms of the length \(L\) of the solenoid:
\[
n = \frac{N}{L}
\]
Where \(N\) is the total number of turns.
### Step 5: Substitute \(n\) into the magnetic moment equation.
We can express the magnetic moment in terms of the flux:
\[
m = \frac{\Phi \cdot L}{\mu_0}
\]
### Step 6: Substitute the values into the equation.
Given:
- \(\Phi = 1.57 \times 10^{-6} \, \text{Wb}\)
- \(L = 60 \, \text{cm} = 0.6 \, \text{m}\)
- \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\)
Now substitute these values into the equation for magnetic moment:
\[
m = \frac{1.57 \times 10^{-6} \cdot 0.6}{4\pi \times 10^{-7}}
\]
### Step 7: Calculate the magnetic moment.
Calculating the above expression:
\[
m = \frac{1.57 \times 10^{-6} \cdot 0.6}{4\pi \times 10^{-7}} \approx \frac{9.42 \times 10^{-7}}{1.25664 \times 10^{-6}} \approx 0.75 \, \text{A m}^2
\]
### Final Answer:
The magnetic moment \(m\) is approximately \(0.75 \, \text{A m}^2\).
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