Skip to content
Question 32 of 58

Q.In a compact coil of 50 turns, the current strength is 10A and the radius of the coil is 25×10⁻² meter. Find the magnitude of the magnetic field at its centre.

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2016Subjective· 2mImportance★★★★★
55% · 32/58 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 →

The magnetic field at the centre of a circular coil of N turns is B = μ₀NI/(2r); substituting the given values gives about 1.26 × 10⁻³ T.

For a single circular loop carrying current I, the magnetic field at its centre (from the Biot-Savart law) is B = μ₀I/(2r). For a coil of N identical turns, the fields of all turns add up: B = μ₀NI/(2r).

Given N = 50, I = 10 A, r = 25 × 10⁻² m = 0.25 m, μ₀ = 4π × 10⁻⁷ T·m/A:

…

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.