A long solenoid carrying a current produces a magnetic field B along its axis. If the number of turns per cm is doubled and the current is made (1/3) rd then the new value of the magnetic field will be (a) B/3 (b) 3B (c) 2B (d) 2B/3

Answer: (d). B = μ₀ n I ∝ nI. B' = μ₀ (2n)(I/3) = 2/3B. So the answer is (d) 2B/3.

A long solenoid carrying a current produces a magnetic field B along… | MHT CET 2021 (21 Sep, Shift 1)

Q38MHT CET 2021 · 21 SEP · SHIFT 1Easy

Ampere's Law, Cylinders, Solenoid & Toroid

A long solenoid carrying a current produces a magnetic field B along its axis. If the number of turns per $\text{cm}$ is doubled and the current is made $\left(\frac{1}{3}\right)$ rd then the new value of the magnetic field will be

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Answer: (d)

$B = \mu_0 n I \propto nI$ .
$B' = \mu_0 (2n)\left(\dfrac{I}{3}\right) = \dfrac{2}{3}B$ .
So the answer is (d) $\dfrac{2B}{3}$ .

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