ExamBro
ExamBro
COMEDK · Physics · 22. Electromagnetic Induction

A vertical circular coil of radius \(0.1 \mathrm{~m}\) and having 10 turns carries a steady current. When the plane of the coil is normal to the magnetic meridian, a neutral point is observed at the centre of the coil. If \(B_{H}=0.314 \times 10^{-4} \mathrm{~T}\), then the current in the coil is

  1. A \(2 \mathrm{~A}\)
  2. B \(1 \mathrm{~A}\)
  3. C \(0.5 \mathrm{~A}\)
  4. D \(0.25 \mathrm{~A}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(0.5 \mathrm{~A}\)

Step-by-step Solution

Detailed explanation

Given, radius of the coil \(=0.1 \mathrm{~m}\) No. of turns in the coil \(=10\) Horizontal component of magnetic field, \[ B_{H}=0.314 \times 10^{-4} \mathrm{~T} \] The magnetic field at the centre of current carrying coil is given by the formula, \(B=\frac{\mu_{0} n I}{2 R}\)…