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MHT CET · Physics · Magnetic Effects of Current

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

  1. A \(\mathrm{B}_1\)
  2. B \(\frac{\mathrm{B}_1}{5}\)
  3. C \(5 \mathrm{~B}_1\)
  4. D \(0.25 \mathrm{~B}_1\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\mathrm{B}_1\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \mathrm{B}_1=\mu_0 \mathrm{n}_1 \mathrm{I}_1 \text { and } \mathrm{B}_2=\mu_0 \mathrm{n}_2 \mathrm{I}_2 \\ & \mathrm{n}_2=5 \mathrm{n}_1 \quad \mathrm{I}_2=0.2 \mathrm{I}_1 \\ & \therefore \frac{\mathrm{B}_2}{\mathrm{~B}_1}=\frac{\mathrm{n}_2}{\mathrm{n}_1} \cdot \frac{\mathrm{I}_2}{\mathrm{I}_1}=5 \times 0.2=1 \\ & \therefore \mathrm{B}_2=\mathrm{B}_1\end{aligned}\)