An emf of 22.0 mV is induced in a 519-turn coil when the current is changing at the rate of 10.0 A/s. What is the magnetic flux through each turn of the coil at an instant when …

Question

asked May 13, 2021 141k views

1 Answer

Alex Stybaev · Answer 1 · 2021-05-18T13:37:32+0000

Answer:

$\phi=1.56* 10^(-5)\ Wb$

Step-by-step 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,

$L=(\epsilon)/((dI/dt))\\\\L=(0.022)/(10)\\\\L=0.0022\ H$

We have,

$L=(N\phi)/(I)$ ,
$\phi$ is magnetic flux

$\phi=(LI)/(N)\\\\\phi=(0.0022*3.7)/(519)\\\\\phi=1.56* 10^(-5)\ Wb$

So, the magnetic flux is
$1.56* 10^(-5)\ Wb$ .