Final answer:
Doubling the current and halving the number of turns per unit length in a solenoid will result in the magnetic field inside the solenoid remaining the same, which is B.
Step-by-step explanation:
The question asks about how the magnetic field inside a solenoid changes if the current is doubled and the number of turns per cm is halved.
The magnetic field B inside a current-carrying 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.
By doubling the current (2I) and halving the number of turns per unit length (n/2), the new magnetic field value would be given by
B' = μ_0 (n/2) (2I)
= μ_0 n I
= B.
Therefore, the new value of the magnetic field will be the same as the original value, that is B.