Given:
The number of the coils, N=20
The cross-sectional area of the solenoid, A=4.0×10⁻⁴ m²
The magnetic fields, B₁=0.0500 T
and B₂=0.180 T
The time interval, t=1.90 s
The resistance of the resistor, R=1.90 s
To find:
1. The magnitude of the induced emf.
2. The current flowing through the resistor.
Explanation:
1.
The magnitude of the induced emf in the solenoid is given by,
[tex]E=NA\frac{(B_2-B_1)}{t}[/tex]On substituting the known values,
[tex]\begin{gathered} E=20\times4.0\times10^{-4}\times\frac{(0.180-0.0500)}{1.90} \\ =0.55\times10^{-3}\text{ V} \\ =0.55\text{ mV} \end{gathered}[/tex]2.
From Ohm's law, The induced emf is given by,
[tex]E=IR[/tex]On substituting the known values,
[tex]\begin{gathered} 0.55\times10^{-3}=I\times5 \\ \Rightarrow I=\frac{0.55\times10^{-3}}{5} \\ =0.11\times10^{-3}\text{ A} \\ =0.11\text{ mA} \end{gathered}[/tex]Final answer:
1. The induced emf is 0.55 mV
2. The current through the given resistor is 0.11 mA
