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

The magnetic field inside a long solenoid is given by:

\[ B = \mu_0 n I \]

where:

•  \(\mu_0\) = \(4\pi \times 10^{-7}\) \(\text{Tm/A}\) (permeability of free space),
• \( n = \frac{70}{\text{cm}} = 70 \times 10^2 \, \text{turns/m}, \)
• \( I = 2.0 \, \text{A} \) (current).

Substitute the values:

\[ B = (4\pi \times 10^{-7}) \cdot (70 \times 10^2) \cdot 2 \]

\[ B = 176 \times 10^{-4} \, \text{T} \]

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