A circular coil (800 turns, radius = 0.063 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 = 0.016 s, the normal makes an angle of 45o with the field because the coil has made one-eighth of a revolution. An average emf of magnitude 0.042 V is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.

Question

Mckenna Hull

Physics

2 December, 09:05

A circular coil (800 turns, radius = 0.063 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 = 0.016 s, the normal makes an angle of 45o with the field because the coil has made one-eighth of a revolution. An average emf of magnitude 0.042 V is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.

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Answers (1)

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Glenn Walls · Answer 1

0.000099T

Explanation:

For a circular coil the flux through a single turn changes by:

∆ = BAcos45° - BAcos90°

= BA cos 45°

During the interval of ∆t = 0.0165s

For N turn, Faraday's law gives the magnitude of emf as follows:

/E / = / - N BAcos45° / : ∆t

Since the loops are circular the area A of each loop is equal to πr square

B = / E / ∆t : Nπr square cos45°

= 0.042V * 0.0165 / 800 * 3.142 * 0.063^2 cos 45°

= 0.000693/7

= 0.000099T

= 9.9*10^-5T