Practice portal › Moving Charges and Magnetism › Ampere's Circuital Law, Solenoid and Toroid

A solenoid of length 0.25 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 2.5 A. What is the magnitude of the magnetic field inside the solenoid? (μ₀=4π×10⁻⁷ SI)

Asked in GUJCET 2022 · Solenoid, toroid and field energy

Answer: (1) 6.28×10⁻³ T

Step-by-step solution

Given: L=0.25 m, N=500 turns, I=2.5 A, μ₀=4π×10⁻⁷ SI. The radius 1 cm is not needed: inside a long solenoid the field does not depend on it.

Idea: B=μ₀ n I, where n is the number of turns per metre.

n=N/L=(500)/(0.25)=2000 m⁻¹.

B=4π×10⁻⁷×2000×2.5=4π×10⁻⁷×5000.

B=2π×10⁻³=6.28×10⁻³ T.

Why the other options are wrong

More Ampere's Circuital Law, Solenoid and Toroid questionsAll Ampere's Circuital Law, Solenoid and Toroid questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer