Question

When the current in a toroidal solenoid is changing at a rate of 0.0200 A/s, the magnitude of the induced emf is 12.7 mV. When the current equals 1.50 A, the average flux through each turn of the solenoid is 0.00458 Wb. How many turns does the solenoid have?

Answer 1

`N = 208 \ turns`

Step-by-step explanation:

From the question we are told that

The rate of current change is
`(di )/(dt) = 0.0200 \ A/s`

The magnitude of the induced emf is
`\epsilon = 12.7 \ mV = 12.7 *10^(-3) \ V`

The current is
`I = 1.50 \ A`

The average flux is
`\phi = 0.00458 \ Wb`

Generally the number of turns the number of turn the solenoid has is mathematically represented as

`N = (\epsilon_o * I)/( \phi * (di)/(dt) )`

substituting values

`N = ( 12.7*10^(-3) * 1.50 )/( 0.00458 * 0.0200 )`

`N = 208 \ turns`