Answer:
[tex]\phi=1.56\times 10^{-5}\ Wb[/tex]
Explanation:
Given that,
Emf, V = 22 mV
Number of turns in the coil us 519
Rate of change of current is 10 A/s.
We need to find the magnetic flux through each turn of the coil at an instant when the current is 3.70 A.
Let's find the inductance first. So,
[tex]L=\dfrac{\epsilon}{(dI/dt)}\\\\L=\dfrac{0.022}{10}\\\\L=0.0022\ H[/tex]
We have,
[tex]L=\dfrac{N\phi}{I}[/tex], [tex]\phi[/tex] is magnetic flux
[tex]\phi=\dfrac{LI}{N}\\\\\phi=\dfrac{0.0022\times3.7}{519}\\\\\phi=1.56\times 10^{-5}\ Wb[/tex]
So, the magnetic flux is [tex]1.56\times 10^{-5}\ Wb[/tex].
