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

Answer:

The number of turns in the solenoid is 230.

Explanation:

Given that,

Rate of change of current, $\dfrac{dI}{dt}=0.0240\ A/s$

Induced emf, $\epsilon=12.4\ mV=12.4\times 10^{-3}\ V$

Current, I = 1.5 A

Magnetic flux, $\phi=0.00338\ Wb$

The induced emf through the solenoid is given by :

$\epsilon=L\dfrac{dI}{dt}$

or

$L=\dfrac{\epsilon}{(di/dt)}$........(1)

The self inductance of the solenoid is given by :

$L=\dfrac{N\phi}{I}$.........(2)

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

$\dfrac{\epsilon}{(di/dt)}=\dfrac{N\phi}{I}$

N is the number of turns in the solenoid

$N=\dfrac{\epsilon I}{\phi (dI/dt)}$

$N=\dfrac{12.4\times 10^{-3}\times 1.5}{0.00338 \times 0.024}$

N = 229.28 turns

or

N = 230 turns

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