A long solenoid carrying current I₁ produces magnetic field B₁ along its axis. If the current is reduced to 20% and number of turns per cm are increased five times then new magnetic field B ₂ is equal to (a) B₁ (b) B₁/5 (c) 5B₁ (d) 0.25B₁

Answer: (a). B = μ₀ n I ∝ nI. Current reduced to 20%: I₂ = 0.2 I₁; n₂ = 5 n₁. B₂ = B₁ × 5 × 0.2 = B₁. So the answer is (a) B₁.

A long solenoid carrying current I₁ produces magnetic field B₁ along… | MHT CET 2021 (23 Sep, Shift 2)

Q37MHT CET 2021 · 23 SEP · SHIFT 2Easy

Ampere's Law, Cylinders, Solenoid & Toroid

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 $\text{cm}$ are increased five times then new magnetic field B ${}_2$ is equal to

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Answer: (a)

$B = \mu_0 n I \propto nI$ .
Current reduced to $20\%$ : $I_2 = 0.2 I_1$ ; $n_2 = 5 n_1$ .
$B_2 = B_1 \times 5 \times 0.2 = B_1$ .
So the answer is (a) $B_1$ .

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Ampere's Law, Cylinders, Solenoid & Toroid question from MHT CET 2021 (23 Sep, Shift 2)

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