Magnetic Field Inside Solenoid Calculation

This problem involves calculating the magnetic field magnitude inside a solenoid given its length, number of turns, and the current flowing through it. We will use the standard formula for the magnetic field inside a long solenoid.

Understanding the Solenoid Magnetic Field Formula

The magnetic field ($B$) inside a long solenoid is uniform and its magnitude is given by the formula:

$B = \mu_0 n I$

Where:

• $B$ is the magnetic field strength (in Tesla, T).
• $\mu_0$ is the permeability of free space, a constant value of \( 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} \).
• $n$ is the number of turns per unit length (in turns/meter).
• $I$ is the current flowing through the solenoid (in Amperes, A).

Calculating Turns Per Unit Length (n)

The number of turns per unit length ($n$) is calculated by dividing the total number of turns ($N$) by the length of the solenoid ($L$).

Given:

• Total number of turns, $N = 800$
• Length of the solenoid, $L = 0.3 \, \text{m}

Therefore, $n$ is:

$n = \frac{N}{L} = \frac{800}{0.3} \, \text{turns/m}$

$n = \frac{8000}{3} \, \text{turns/m} \approx 2666.67 \, \text{turns/m}$

Calculating the Magnetic Field (B)

Now, we can substitute the values of $\mu_0$, $n$, and $I$ into the magnetic field formula.

Given:

• $\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$
• $n \approx 2666.67 \, \text{turns/m}$
• Current, $I = 6 \, \text{A}$

Substituting these values:

$B = (4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}) \times \left(\frac{800}{0.3} \, \text{turns/m}\right) \times (6 \, \text{A})$

$B = (4\pi \times 10^{-7}) \times \left(\frac{800}{0.3}\right) \times 6 \, \text{T}$

$B = (4\pi \times 10^{-7}) \times (2666.67) \times 6 \, \text{T}$

$B = (4\pi \times 10^{-7}) \times 16000 \, \text{T}$

$B \approx (12.566 \times 10^{-7}) \times 16000 \, \text{T}$

$B \approx 0.0201056 \, \text{T}$

Converting to Millitesla (mT)

The options are given in milliTesla (mT). To convert Tesla to milliTesla, we multiply by 1000.

$B \, (\text{in mT}) = B \, (\text{in T}) \times 1000$

$B \approx 0.0201056 \times 1000 \, \text{mT}$

$B \approx 20.1056 \, \text{mT}$

Comparing with Options

The calculated value of the magnetic field is approximately $20.1$ mT. Comparing this with the given options:

• 1. 2.03 T
• 2. 60.3 mT
• 3. 20 mT
• 4. 6.03 T

The calculated value $20.1$ mT is closest to the option $20$ mT.

Final Answer Determination

The magnitude of the magnetic field inside the solenoid is approximately $20.1$ mT, which closely matches option 3.