alecnewman2002
07.04.2020 •
Physics
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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Ответ:
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
Ответ:
The answer is 15.