Question

A circular coil (670 turns, radius = 0.079 m) is rotating in a uniform magnetic field. At t = 0 s, the normal to the coi...


A circular coil (670 turns, radius = 0.079 m) is rotating in a uniform magnetic field. At t = 0 s, the normal to the coil is perpendicular to the magnetic field. At t = 0.020 s, the normal makes an angle of 45 degrees with the field because the coil has made one-eighth of a revolution. An average emf of magnitude 0.052 V is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.

Answer

Given data:

  • Number of turns is N=670.
  • Radius of coil is r=0.079 m.
  • Angle made at t=0.020 s is θ=45°.
  • Emf induced is ε=0.052 V.

The magnetic flux is given by,

ϕ=BAcosθϕ=Bπr2cosθϕ=Bπ0.079 m2cos45°ϕ=13.86×10-3 m2B

The induced emf in the coil is,

ε=-Nϕtε=Nϕt

Substitute all value in above expression,

0.052 V=670×13.86×10-3 m2B0.020 sB=1.12×10-4 V·s/m2×1 T1 V·s/m2B=1.12×10-4 T

Thus, the magnitude of magnetic field is 1.12×10-4 T.

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