NEET2026PhysicsChapterActual
A circular coil of 50 turns and radius 10 cm has a magnetic dipole moment of 1.57 A m ^2 . The current flowing through the coil and the magnitude of the magnetic field at its centre are, respectively: (Take = 3.14 and ₀ = 4 10⁻⁷ T m/A )
Options
- A1 A , 6.28 10⁻⁴ T
- B50 A , 1.57 10⁻² T
- C1 A , 3.14 10⁻⁴ T
- D1 A , 6.28 10⁻⁶ T
Correct answer
C. 1 A , 3.14 10⁻⁴ T
Step-by-step solution
Given N = 50 , r = 0.1 m , and M = 1.57 A m ^2 . The magnetic dipole moment of a coil is given by M = N I A = N I ( r^2) . Substituting the given values: 1.57 = 50 I 3.14 (0.1)^2 1.57 = 50 I 0.0314 1.57 = 1.57 I I = 1 A The magnetic field at the centre of the circular coil is given by B = ₀ N I 2 r . Substituting the values: B = 4 10⁻⁷ 50 1 2 0.1 B = 200 10⁻⁷ 0.2 B = 1000 10⁻⁷ B = 3.14 10⁻⁴ T Answer: 1 A , 3.14 10⁻⁴ T