Respuesta :
Answer:
A)[tex]\Phi=83.84\times 10^{-9} [/tex]
B)[tex]\Phi=0 Wb[/tex]
C)[tex]emf=5.4090\times 10^{-4}V [/tex]
Explanation:
Given that:
- no. of turns i the coil, [tex]n=200[/tex]
- area of the coil, [tex]a=13.1 \times 10^{-4}\,m^2[/tex]
- time interval of rotation, [tex]t=3.1\times 10^{-2}\,s[/tex]
- intensity of magnetic field, [tex]B=6.4\times 10^{-5}\,T[/tex]
(A)
Initially the coil area is perpendicular to the magnetic field.
So, magnetic flux is given as:
[tex]\Phi=B.a\,cos \theta[/tex]..................................(1)
[tex]\theta[/tex] is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.
[tex]\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ} [/tex]
[tex]\Phi=83.84\times 10^{-9} Wb[/tex]
(B)
In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.
∴ [tex]\theta=90^{\circ}[/tex]
From eq. (1)
[tex]\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ} [/tex]
[tex]\Phi=0 Wb[/tex]
(C)
According to the Faraday's Law we have:
[tex]emf=n\frac{B.a}{t}[/tex]
[tex]emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}[/tex]
[tex]emf=5.4090\times 10^{-4}V [/tex]
