A long solenoid carrying a current produces a magnetic field $B$ along its axis. If the current is doubled…
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 the magnetic field is:
$\frac{B}{2}$
$B$
$2 B$
$4 B$
Solution
Magnetic field induction at point insides the solenoid of length $l$, having $n$ turns per unit length carrying current $i$. is given by
$B=\mu_0 n i$
If $i$ is double and $n$ is halved then $B$ remains same.