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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:
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The correct answer is:
$B$
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.
$$
B=\mu_0 n i
$$
If $i$ is double and $n$ is halved then $B$ remains same.
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