A circular coil (910 turns, radius = 0.077 m) is rotating in a uniform magnetic
ID: 1465100 • Letter: A
Question
A circular coil (910 turns, radius = 0.077 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 0.065 V is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.A circular coil (910 turns, radius = 0.077 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 0.065 V is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.
Explanation / Answer
No of turns, N = 910
area of the coil, A = pi*r^2
= pi*0.077^2
= 0.0186 m^2
Let B is the magnetic field.
Initial magnetic flux through the coil = N*A*B*cos(90)
= 0
final magnetic flux = N*A*B*cos(45)
we know,
induced emf = change in flux/time
emf = N*A*B*cos(45)/t
==> B = emf*t/(N*A*cos(45))
= 0.065*0.016/(910*0.0186*cos(45))
= 8.69*10^-5 T <<<<<<-----------------Answer
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