EXAMPLE 8.1: MAGNETIC FIELD OF A SOLENOID A student takes a long straight copper wire (with insulation on it) from his physics is the wire on an iron rod of 50 cm , to make it solenoid of 15 , turns. He connects the en wire with the battery, which provides 1.2 A current. Find the magnetic fleld he solenoid by doing so. GIVEN Length ' $$ L $$ ' $$ =50 \mathrm{~cm}=0.5 \mathrm{~m} $$ Number of turns ' N ' $$ =50 $$ Eurrent $$ { }^{\prime \prime} $$ ' $$ =1.2 \mathrm{~A} $$ REQUIRED Magnetic Field Strength ' $$ B $$ '

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We use the solenoid formula for the magnetic field inside a long solenoid:

B = μ₀ (N/L) I

where
 • μ₀ = 4π×10⁻⁷ T·m/A (the permeability of free space),
 • N is the number of turns,
 • L is the length of the solenoid in meters, and
 • I is the current in amperes.

Given:
 L = 50 cm = 0.5 m
 N = 50 turns
 I = 1.2 A

Step 1. Find the number of turns per unit length (n):
  n = N / L = 50 / 0.5 = 100 turns per meter

Step 2. Substitute into the formula for B:
  B = μ₀ n I = (4π×10⁻⁷ T·m/A) × (100/m) × (1.2 A)

Step 3. Compute the product:
  B = 4π×10⁻⁷ × 120 = 480π×10⁻⁷ T

Evaluating numerically (using π ≈ 3.14):
  B ≈ 480×3.14×10⁻⁷ T ≈ 1507×10⁻⁷ T ≈ 1.51×10⁻⁴ T

Thus, the magnetic field strength inside the solenoid is approximately 1.51×10⁻⁴ tesla.

Note:
Although the student wound the wire around an iron rod, the calculation above uses μ₀ (as for an air core). In a real situation an iron core would increase the field by a factor equal to the iron’s relative permeability (μr). However, since no μr was given, we have taken the core to contribute only the free‐space permeability.
The magnetic field strength inside the solenoid is approximately 1.51×10⁻⁴ tesla.

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