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
Answer & explanation
Correct answer: option 4
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: $B$