Question

A solenoid of 2.5 m length and 2.0 cm diameter possesses 10 turns per cm. A current of 0.5 A is flowing through it. The magnetic induction at axis inside the solenoid is:

• A. $2π×10^-4T$
• B. $2π×10^-5T$
• C. $2π×10^-6T$
• D. $2π×10^-7T$

Hint:

Magnetic field inside a long solenoid: $B = \mu₀ n I$.

Answer 1

The Correct Option is A

To determine the magnetic induction at the axis inside the solenoid, we will use the formula for the magnetic field inside a long solenoid. The formula is given by:

\(B = \mu_0 n I\)

Where:

• \(B\) is the magnetic field.
• \(\mu_0\) is the permeability of free space, which is \(4\pi \times 10^{-7} \, T \, m/A\).
• \(n\) is the number of turns per unit length (turns per meter).
• \(I\) is the current through the solenoid.

Let's calculate step-by-step:

1. Convert the number of turns per cm to turns per meter. Since the solenoid has 10 turns per cm, in terms of meters, it will be:

\(n = 10 \, \text{turns/cm} = 1000 \, \text{turns/m}\)

1. Substitute the known values into the formula:

\(B = \mu_0 n I = (4\pi \times 10^{-7} \, T \, m/A) \times (1000 \, \text{turns/m}) \times (0.5 \, A)\)

1. Calculate the magnetic field:

\(B = 2 \pi \times 10^{-4} \, T\)

The magnetic induction at the axis inside the solenoid is therefore \(2\pi \times 10^{-4} \, T\).

Thus, the correct answer is \(2\pi \times 10^{-4} \, T\), which confirms that the selection of the correct option is accurate.