Question

An emf of 17.0 mV is induced in a 483-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 the current is 4.30 A?

Answer 1

`0.0000151Tm^(2)`

Step-by-step explanation:

`E=17.0mV=0.017V\\t=10s\\I=4.30A`

`N=483 turns`

The relation to find the magnetic flux is:

φ
`=(LI)/(N)`

so we need to find L

`L=(E)/(dI/dt)`

`L=(0.017)/(10.0) \\L=0.0017H`

Now we can apply:

φ=
`(L*I)/(N)`

φ=
`(0.0017*4.30)/(483)`

φ=
`0.0000151Tm^(2)`