Question

A long solenoid is formed by winding 70 turns/cm. If 2.0 A current flows, then the magnetic field produced inside the solenoid is:

• A. \(88 \times 10^{-4} T\)
• B. \(123.2 \times 10^{-4} T\)
• C. \(352 \times 10^{-4} T\)
• D. \(176 \times 10^{-4} T\)

Hint:

The formula \( B = \mu_0 n I \) is key for solenoids. Always convert \( n \) into turns/m (SI unit) before substituting.

Answer 1

The Correct Option is D

To find the magnetic field inside a solenoid, we use the formula for the magnetic field \( B \) inside a long solenoid:

\(B = \mu_0 \, n \, I\)

where:

• \(\mu_0\) is the permeability of free space, having a value of \(4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\).
• \(n\) is the number of turns per unit length (in turns per meter).
• \(I\) is the current in amperes.

Given: \(n = 70 \, \text{turns/cm} = 70 \times 10^2 \, \text{turns/m}\) and \(I = 2.0 \, \text{A}\).

First, convert the number of turns per centimeter to turns per meter:

\(n = 70 \times 100 = 7000 \, \text{turns/m}\)

Now, substitute the values into the formula:

\(B = 4\pi \times 10^{-7} \times 7000 \times 2.0\)

Calculate the result:

\(B = 4\pi \times 10^{-7} \times 14000\)

\(B = 56\pi \times 10^{-4} \, \text{T}\)

Approximating \(\pi \approx 3.14\),

\(B \approx 56 \times 3.14 \times 10^{-4} \, \text{T}\)

\(B \approx 176 \times 10^{-4} \, \text{T}\)

Therefore, the magnetic field produced inside the solenoid is \(176 \times 10^{-4} \, \text{T}\).

Thus, the correct answer is:

\(176 \times 10^{-4} \, \text{T}\)