Final answer:
The ratio of the magnetic flux densities at the axis of a solenoid when it's original length is 30 cm and after removing 5 cm from each side (resulting in 20 cm length) is 6:5.
Step-by-step explanation:
The magnetic field B inside a solenoid is given by the formula B = μ_0 * (n * I), where μ_0 is the permeability of free space, n is the number of turns per unit length, and I is the current. For a solenoid of length l with negligible end effects, the number of turns N is directly proportional to its length, thus n = N / l. In the given problem, the original length of the solenoid is 30 cm, and after removing 5 cm from each side, the length becomes 20 cm. This results in a corresponding decrease in the number of turns and hence a decrease in n.
The ratio of the magnetic field before and after the length change can be calculated as follows: B1/B2 = n1/n2. Initially, n1 is based on the full 30 cm length, and after the change, n2 is based on the 20 cm length. Since the lengths and number of turns are directly proportional, this yields a ratio of the magnetic flux densities at their axis as 6:5 (option c).