To find the magnitude of the induced electric field at points near the center of a long thin solenoid, we can use Faraday's law of electromagnetic induction. This law states that a changing magnetic field within a closed loop induces an electromotive force (EMF) in that loop. For a solenoid, the magnetic field inside it is uniform and can be calculated using the formula:
Magnetic Field Inside the Solenoid
The magnetic field \( B \) inside a solenoid is given by the equation:
B = \mu_0 n I
where:
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- n is the number of turns per unit length (in turns/meter), and
- I is the current through the solenoid.
In this case, the solenoid has 900 turns/m and a radius of 2.50 cm, with a current increasing at a rate of 60 A/s. First, we need to find the rate of change of the magnetic field \( \frac{dB}{dt} \).
Calculating the Rate of Change of Magnetic Field
Since the current is changing, we can express the change in magnetic field over time as:
\( \frac{dB}{dt} = \mu_0 n \frac{dI}{dt} \)
Substituting the values:
- \( n = 900 \, \text{turns/m} \)
- \( \frac{dI}{dt} = 60 \, \text{A/s} \)
Now, substituting these into the equation:
\( \frac{dB}{dt} = (4\pi \times 10^{-7}) \times 900 \times 60 \)
Calculating this gives:
\( \frac{dB}{dt} \approx 0.000678 \, \text{T/s} \)
Induced Electric Field Calculation
According to Faraday's law, the induced electric field \( E \) at a distance \( r \) from the axis of the solenoid can be calculated using the formula:
\( E = -\frac{1}{2\pi r} \frac{d\Phi_B}{dt} \)
Where \( \Phi_B \) is the magnetic flux through a circular loop of radius \( r \). The magnetic flux is given by:
\( \Phi_B = B \cdot A \)
For a circular area \( A = \pi r^2 \), we have:
\( \frac{d\Phi_B}{dt} = A \frac{dB}{dt} = \pi r^2 \frac{dB}{dt} \)
Calculating for Different Distances
Now, we can calculate the induced electric field at the two specified distances from the axis of the solenoid:
1. At 0.5 cm from the axis
Substituting \( r = 0.005 \, \text{m} \):
\( E = -\frac{1}{2\pi (0.005)} \cdot \pi (0.005)^2 \cdot \frac{dB}{dt} \)
After simplification, this becomes:
\( E = -\frac{(0.005)}{2} \cdot \frac{dB}{dt} \)
Now substituting \( \frac{dB}{dt} \):
\( E \approx -\frac{(0.005)}{2} \cdot 0.000678 \approx -0.00001695 \, \text{V/m} \)
2. At 1.0 cm from the axis
Substituting \( r = 0.01 \, \text{m} \):
\( E = -\frac{1}{2\pi (0.01)} \cdot \pi (0.01)^2 \cdot \frac{dB}{dt} \)
After simplification:
\( E \approx -\frac{(0.01)}{2} \cdot \frac{dB}{dt} \)
Substituting \( \frac{dB}{dt} \):
\( E \approx -\frac{(0.01)}{2} \cdot 0.000678 \approx -0.0000339 \, \text{V/m} \)
Final Results
To summarize, the magnitudes of the induced electric fields are:
- At 0.5 cm from the axis: approximately 0.00001695 V/m
- At 1.0 cm from the axis: approximately 0.0000339 V/m
This demonstrates how the induced electric field varies with distance from the solenoid's axis, reflecting the relationship between the changing magnetic field and the induced electric field in accordance with Faraday's law.
