A long solenoid carrying a current produces a magnetic field $$B$$ along its axis. If the current is doubled and the number of turns per cm is halved, the new value of magnetic field will be equal to

The magnetic field inside a long solenoid is given by:

$$B = \mu_0 n I$$

where $$n$$ is the number of turns per unit length and $$I$$ is the current.

Initially, the magnetic field is:

$$B = \mu_0 n I$$

When the current is doubled ($$I' = 2I$$) and the number of turns per cm is halved ($$n' = \frac{n}{2}$$):

$$B' = \mu_0 n' I' = \mu_0 \times \frac{n}{2} \times 2I = \mu_0 n I = B$$

The new magnetic field equals the original magnetic field.

Hence, the correct answer is Option D.