The current in an ideal, long solenoid is changing at a steady rate of 0.01 A s ⁻¹ . The solenoid features 2000 turns/m and a radius of 6.0 cm . Consider a circular loop of radius 1.0 cm located inside the solenoid, with its axis matching the solenoid's axis. Determine the change in the magnetic flux through this loop over a time interval of 2.0 seconds .

Step-by-step solution

The magnetic field inside a long solenoid is given by B = ₀ n i The magnetic flux through the small circular loop is = B A = ₀ n i ( r^2) The change in magnetic flux over a time interval t is = ₀ n ( r^2) i The change in current over t = 2.0 s is i = ( di dt ) t = 0.01 2.0 = 0.02 A Substituting the given values: = (4 10⁻⁷) 2000 (0.01)^2 0.02 = 16 ^2 10⁻¹⁰ Using ^2 9.87 , we get: 157.9 10⁻¹⁰ 1.6 10⁻⁸ Wb