Skip to content
Question 29 of 46

Q.State and explain Ampere's circuital law.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2018Subjective· 2mImportance★★★★★
63% · 29/46 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Ampere's circuital law relates the circulation of the magnetic field around any closed path to the total current threading that path, and lets BB be found easily in symmetric situations.

Statement: The line integral of the magnetic field B⃗\vec B around any closed path (Amperian loop) is equal to μ0\mu_0 times the total current enclosed by that path:

∮B⃗⋅dl⃗=μ0Ienc.\oint \vec B\cdot d\vec l = \mu_0 I_{\text{enc}}.

Explanation with an example: Consider a long straight current-carrying conductor carrying current II, and choose a circular Amperian loop of radius rr centred on the wire, lying in a plane perpendicular to it. By symmetry, B⃗\vec B has the same magnitude at every point on this loop and is everywhere tangential to it (directed along the loop, per the right-hand rule), so B⃗⋅dl⃗=B dl\vec B\cdot d\vec l = B\,dl everywhere. Then

∮B⃗⋅dl⃗=B∮dl=B(2πr).\oint \vec B\cdot d\vec l = B\oint dl = B(2\pi r).

Since the loop encloses the entire current II, Ampere's law gives B(2πr)=μ0IB(2\pi r) = \mu_0 I, so

B=μ0I2πr,B = \frac{\mu_0 I}{2\pi r}, …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.