The magnetic flux near the axis and inside the air core solenoid of length 60 cm carrying current I is 2 10⁻⁶ Wb. Its magnetic moment will be (cross-sectional area is very small as compared to length of solenoid, ₀ = Permeability of free space = 4 10⁻⁷ SI unit)

Step-by-step solution

The magnetic field inside a long solenoid is given by B = ₀ n I = ₀ N L I . The magnetic flux through the cross-section of the solenoid is = B A = ₀ N I A L . The magnetic moment of the solenoid is M = N I A . Substituting M into the flux equation, we get = ₀ M L . Rearranging for M , we have M = L ₀ . Given = 2 10⁻⁶ Wb , L = 60 cm = 0.6 m , and ₀ = 4 10⁻⁷ T m/A . M = 2 10⁻⁶ 0.6 4 10⁻⁷ M = 0.3 10⁻⁶ 4 10⁻⁷ M = 0.3 4 10 = 3 4 = 0.75 Am ^2 Answer: 0.75 Am ^2