Practice portal › Moving Charges and Magnetism › Ampere's Circuital Law, Solenoid and Toroid
Asked in AIPMT 1999 · Thick wires and cables
Idea: apply Ampere's law on a circle inside the cavity and on one outside the pipe.
Inside the hollow, r<rᵢₙₙₑᵣ: the circle encloses no current, so B(2π r)=μ₀×0.
That gives B=0 everywhere inside the cavity.
Outside the pipe, r>rₒᵤₜₑᵣ: the circle encloses the whole current I, so B(2π r)=μ₀I.
B=(μ₀I)/(2π r) — the same as for a solid wire carrying I.
So the field exists outside the pipe only.
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