Question

When the current in a toroidal solenoid is changing at a rate of 0.0240 A/s , the magnitude of the induced emf is 12.4 mV . When the current equals 1.50 A , the average flux through each turn of the solenoid is 0.00338 Wb. How many turns does soleniod has?

Answer 1

The number of turns in the solenoid is 230.

Step-by-step explanation:

Given that,

Rate of change of current,
`(dI)/(dt)=0.0240\ A/s`

Induced emf,
`\epsilon=12.4\ mV=12.4* 10^(-3)\ V`

Current, I = 1.5 A

Magnetic flux,
`\phi=0.00338\ Wb`

The induced emf through the solenoid is given by :

`\epsilon=L(dI)/(dt)`

or

`L=(\epsilon)/((di/dt))`........(1)

The self inductance of the solenoid is given by :

`L=(N\phi)/(I)`.........(2)

From equation (1) and (2) we get :

`(\epsilon)/((di/dt))=(N\phi)/(I)`

N is the number of turns in the solenoid

`N=(\epsilon I)/(\phi (dI/dt))`

`N=(12.4* 10^(-3)* 1.5)/(0.00338 * 0.024)`

N = 229.28 turns

or

N = 230 turns

So, the number of turns in the solenoid is 230. Hence, this is the required solution.