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 $\mathrm{cm}$ is halved, the new value of magnetic field will be equal to

Magnetic field due to long solenoid along its axis is given as,

$B={\mu }_{0}nI$, where $n$ is the number of turns in the solenoid per unit length, $I$ is current in the solenoid and ${\mu }_0$ is a constant called the permeability of free space.

$\Rightarrow \frac{B_1}{ B_2}=\frac{n_1{I}_1}{n_2{I}_2}$

According to the question, $n_2=\frac{n_1}{2} \text{and} I_2=2I_1$

$\Rightarrow \frac{B}{B_2}=\frac{n_1}{\left({\displaystyle \frac{n_1}{2}}\right)}\times \frac{I_1}{2I_1}\Rightarrow B_2=B$; means, the magnetic field remains the same.