Question

The flux through the coils of a solenoid changes from 2.57.10-5 Wb to 9.44.10-5 Wb in 0.0154 s. If 4.08 V of EMF is generated, how many loops does the solenoid have? (No unit)​

Answer 1

Hello!

We can use Faraday's Law of Electromagnetic Induction to solve.

$\epsilon = -N \frac{d\Phi_B}{dt}$

ε = Induced emf (4.08 V)
N = Number of loops (?)

$\Phi_B$ = Magnetic Flux (Wb)
t = time (s)

**Note: The negative sign can be disregarded for this situation. The sign simply shows how the induced emf OPPOSES the current.

Now, we know that $\frac{d\Phi_B}{dt}$ is analogous to the change in magnetic flux over change in time, or $\frac{\Delta \Phi_B}{\Delta t}$, so:
$\epsilon = N \frac{\Delta \Phi_B}{\Delta t}\\\\\epsilon = N \frac{\Phi_{Bf} - \Phi_{Bi}}{\Delta t}$

Rearrange the equation to solve for 'N'.

$N = \frac{\epsilon}{ \frac{\Phi_{Bf} - \Phi_{Bi}}{\Delta t}}$

Plug in the given values to solve.

$N = \frac{4.08}{ \frac{9.44*10^{-5} - 2.57*10^{-5}}{0.0154}} = 914.585 = \boxed{915 \text{ coils}}$

**Rounding up because we cannot have a part of a loop.