Respuesta :
Given:
no. of turn in coil is
[tex]n=500[/tex]radius of the coil is
[tex]r=0.06\text{ m}[/tex]rate of change of magnetic field is
[tex]\frac{dB}{dt}=0.160\text{ T/s}[/tex]coil makes an angle with the magnetic field is
[tex]\theta=40^{\circ}[/tex]Required: magnitude of the resulting emf in the coil
Explanation:
emf in the any coil is given by
[tex]\epsilon=-nA\frac{dB}{dt}\cos40^{\circ}[/tex]plugging all the values in the above relation, we get
[tex]\begin{gathered} \epsilon=500\times3.14\times(0.06\text{ m})^2\times0.160\text{ }\times0.766 \\ \epsilon=0.693\text{ V} \end{gathered}[/tex]Thus, the emf in the coil is
[tex]0.693\text{ V}[/tex]