The magnetic field intensity inside current carrying solenoid is $\mathrm{H}=2.4 \times 10^3 \mathrm{~A} /…

The magnetic field intensity inside current carrying solenoid is $\mathrm{H}=2.4 \times 10^3 \mathrm{~A} / \mathrm{m}$. If length and number of turns of a solenoid is 15 cm and 60 turns respectively. The current flowing in the solenoid is
  1. 4 A
  2. 6 A
  3. $\quad 0.6 \mathrm{~A}$
  4. 60 A

Solution

$\begin{aligned} & B=\frac{\mu_0 \mathrm{NI}}{\mathrm{L}} \\ & \mu_0 \mathrm{H}=\frac{\mu_0 \mathrm{NI}}{\mathrm{L}} \\ & I=\frac{\mathrm{HL}}{\mathrm{n}}=\frac{2.4 \times 10^3 \times 15 \times 10^{-2}}{60}=6 \mathrm{~A}\end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 1)

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