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(A) The magnetic field inside a long solenoid is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length and $I$ is the current.Init

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$B_1$

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(A) The magnetic field inside a long solenoid is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length and $I$ is the current.Initially,$B_1 = \mu_0 n_1 I_1$.Given that the current is reduced to $20 \%$,the new current $I_2 = 0.2 I_1$.The number of turns per $cm$ is increased five times,so the new turn density $n_2 = 5 n_1$.The new magnetic field is $B_2 = \mu_0 n_2 I_2$.Substituting the values: $B_2 = \mu_0 (5 n_1) (0.2 I_1) = \mu_0 n_1 I_1 (5 	imes 0.2) = \mu_0 n_1 I_1 (1) = B_1$.Therefore,the new magnetic field $B_2$ is equal to $B_1$.

$A$ long solenoid carrying current $I_1$ produces a magnetic field $B_1$ along its axis. If the current is reduced to $20 \%$ and the number of turns per $cm$ is increased five times,then the new magnetic field $B_2$ is equal to:

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