Solenoid Magnetic Field

Solenoid Magnetic Field Calculation

Question: A 1.0 m solenoid with 4250 turns carries 5.0 A current. Magnetic field at its center is:

(a) 5.4 × 10⁻² Wb/m²

(b) 2.7 × 10⁻² Wb/m²

(c) 1.35 × 10⁻² Wb/m²

(d) 0.675 × 10⁻² Wb/m²

Solution

Step 1: Formula for Solenoid Magnetic Field
\[ B = \mu_0 \cdot n \cdot I \quad \text{where } n = \frac{N}{L} \]

Step 2: Calculate Turns per Unit Length (\(n\))
\[ n = \frac{4250 \, \text{turns}}{1.0 \, \text{m}} = 4250 \, \text{turns/m} \]

Step 3: Substitute Values
\[ B = (4\pi \times 10^{-7}) \cdot 4250 \cdot 5.0 = 4\pi \cdot 21,\!250 \times 10^{-7} \] \[ \approx 2.67 \times 10^{-2} \, \text{T} \quad (\text{using } \pi \approx 3.1416) \]

Answer: Option (b) \( 2.7 \times 10^{-2} \, \text{Wb/m}^2 \) is correct.