Respuesta :
Answer:
[tex]E=0.2909\ V[/tex]
Explanation:
Given:
no. of turns in the coil, [tex]n=80[/tex]
magnetic field, B = 1.3 T
area of the coil, [tex]A=(0.25\times 0.4)=0.1\ m^2[/tex]
initial angle of the coil plane from the magnetic field, [tex]\psi_i=37^{\circ}[/tex]
final angle of the coil plane from the magnetic field, [tex]\psi_f=90^{\circ}[/tex]
time taken to rotate through the angle, [tex]dt=0.09\ s[/tex]
Now we find initial and final flux through the coil:
[tex]\phi_i=B.A\ cos\ \psi[/tex]
[tex]\phi_i=1.3\times 0.1\ cos\ 37^{\circ}[/tex]
[tex]\phi_i=0.1038\ Wb[/tex]
&
[tex]\phi_f=B.A\ cos\ \psi[/tex]
[tex]\phi_f=1.3\times 0.1\ cos\ 90^{\circ}[/tex]
[tex]\phi_f=0.13\ Wb[/tex]
[tex]\therefore d \phi=\phi_f-\phi_i[/tex]
[tex] d \phi=0.13-0.1038[/tex]
[tex] d \phi=0.0262\ Wb[/tex]
Now we know according to Faraday's Law, emf induced in a coil is given by:
[tex]E=n.\frac{d \phi}{dt}[/tex]
[tex]E=80\times \frac{0.0262}{0.09}[/tex]
[tex]E=0.2909\ V[/tex]
