Physics Circle home Student Login

Home›Chapters›Moving Charges and Magnetism PYQs›Ampere's Law, Cylinders, Solenoid & Toroid›Question 23

MHT CET 2024 (9 May, Shift 1) · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from MHT CET 2024 (9 May, Shift 1)

Q23MHT CET 2024 · 9 MAY · SHIFT 1Easy

Ampere's Law, Cylinders, Solenoid & Toroid

The magnetic field intensity inside current carrying solenoid is H=2.4×103 A/m\mathrm{H} = 2.4 \times 10^{3}\text{ A/m}. If length and number of turns of a solenoid is 15 cm and 60 turns respectively. The current flowing in the solenoid is

Show answer and solution

Answer: (b)

  1. H=nIH = nI, with n=600.15=400n = \dfrac{60}{0.15} = 400 turns/m.
  2. I=Hn=2.4×103400=6I = \dfrac{H}{n} = \dfrac{2.4 \times 10^{3}}{400} = 6 A.
  3. So the answer is (b) 6 A.
Report a mistake

More questions on Ampere's Law, Cylinders, Solenoid & Toroid