A long solenoid carrying a current produces a magnetic field B along its axis. If the value of current is halved and the number of turns per unit length is doubled then the new value of magnetic field is

B

The correct answer is Option (4) → B

$B = \mu_0 n I$

$\text{New current } = \frac{I}{2}, \quad \text{New turns per unit length } = 2n$

$B' = \mu_0 (2n)\left(\frac{I}{2}\right)$

$B' = \mu_0 n I$

$B' = B$

Final Answer: $B$