NEET2026PhysicsChapterActual
A tightly wound circular coil has 40 turns, each of radius 10 cm . If it carries a steady current of 4 A , what is the magnitude of the magnetic field on its axis at a distance of 10 3 cm from the centre?
Options
- A4 10⁻⁵ T
- B3.2 10⁻⁴ T
- C8 10⁻⁵ T
- D10⁻⁶ T
Correct answer
A. 4 10⁻⁵ T
Step-by-step solution
The magnetic field on the axis of a circular coil is given by the formula: B = ₀ N I R^2 2(R^2 + x^2)^ 3/2 Given values are: N = 40 R = 10 cm = 0.1 m I = 4 A x = 10 3 cm = 0.1 3 m ₀ = 4 10⁻⁷ T m/A First, calculate the term (R^2 + x^2) : R^2 + x^2 = (0.1)^2 + (0.1 3 )^2 = 0.01 + 0.03 = 0.04 m ^2 Now, calculate (R^2 + x^2)^ 3/2 : (0.04)^ 3/2 = (0.2^2)^ 3/2 = (0.2)^3 = 0.008 m ^3 Substituting these values into the formula for B : B = 4 10⁻⁷ 40 4 (0.1)^2 2 0.008 B = 4 10⁻⁷ 1.6 0.016 B = 4 10⁻⁷ 100 B = 4 10⁻⁵ T Answer: