The magnetic flux near the axis and inside the air core solenoid of length 60 cm carrying current 'I' is 1.57 10⁻⁶ Wb . Its magnetic moment will be cross-sectional area of a solenoid is very small as compared to its length, ₀ = 4 10⁻⁷ SI unit

Step-by-step solution

The magnetic field inside a long solenoid is given by: B = ₀ N I L where N is the total number of turns, I is the current, and L is the length of the solenoid. The magnetic flux through the cross-section of the solenoid is: = B A = ₀ N I A L The magnetic moment of a solenoid is given by: M = N I A Substituting M into the flux equation, we get: = ₀ M L Rearranging to solve for the magnetic moment M : M = L ₀ Given values are: = 1.57 10⁻⁶ Wb L = 60 cm = 0.6 m ₀ = 4 10⁻⁷ T m/A = 4 3.14 10⁻⁷ T m/A = 12.56 10⁻⁷ T m/A Su