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A long solenoid carrying current $\mathrm{I}_1$ produces magnetic field $\mathrm{B}_1$ along its axis. If the current is reduced to $20 \%$ and number of turns per $\mathrm{cm}$ are increased five times, then new magnetic field $\mathrm{B}_2$ is equal to
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$\mathrm{B}_1$
$\begin{aligned} & \mathrm{B}_1=\mu_0 \mathrm{n}_1 \mathrm{I}_1 \text { and } \mathrm{B}_2=\mu_0 \mathrm{n}_2 \mathrm{I}_2 \\ & \mathrm{n}_2=5 \mathrm{n}_1 \quad \mathrm{I}_2=0.2 \mathrm{I}_1 \\ & \therefore \frac{\mathrm{B}_2}{\mathrm{~B}_1}=\frac{\mathrm{n}_2}{\mathrm{n}_1} \cdot \frac{\mathrm{I}_2}{\mathrm{I}_1}=5 \times 0.2=1 \\ & \therefore \mathrm{B}_2=\mathrm{B}_1\end{aligned}$
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