Solenoid Magnetic Induction

Solenoid Magnetic Induction

Question: A solenoid (1.5 m length, 4.0 cm diameter) has 10 turns/cm. With 10 A current, the magnetic field at its axis is:

(a) \( 4\pi \times 10^{-3} \, \text{T} \)

(b) \( 2\pi \times 10^{-5} \, \text{T} \)

(c) \( 2\pi \times 10^{-2} \, \text{G} \)

(d) \( 2\pi \times 10^{-5} \, \text{G} \)

Solution

Step 1: Formula for Solenoid Magnetic Field
\[ B = \mu_0 n I \quad \text{where } n = \frac{\text{total turns}}{\text{length}}. \]

Step 2: Calculate Turns per Meter (\(n\))
Given: 10 turns/cm = \( 10 \times 100 = 1000 \, \text{turns/m} \).

Step 3: Substitute Values
\[ B = (4\pi \times 10^{-7} \, \text{T·m/A}) \cdot (1000 \, \text{turns/m}) \cdot (10 \, \text{A}) \] \[ = 4\pi \times 10^{-3} \, \text{T}. \]

Answer: Option (a) \( 4\pi \times 10^{-3} \, \text{T} \) is correct.