Question

When the current in a toroidal solenoid is changing at a rate of 0.0260 A/s, the magnitude of the induced emf is 12.6 mV. When the current equals 1.40 A, the average flux through each turn of the solenoid is 0.00285 Wb. How many turns does the solenoid have?

Short answer

The number of turns that the solenoid possess are 238 turns.

Step-by-step solution

Given Data

Faraday’s law states that a current is induced in a conductor when it is exposed to a time varying magnetic flux. This induced current is driven by a force called electromotive or electromagnetic force. The magnitude of induced emf is given by

$\epsilon =-L\frac{di}{dt}$

Where L is the inductance of the conductor.

An inductor is a passive two-terminal device that stores energy in a magnetic field when current passes through it.

Lenz further explained the direction of this induced current. According to lens, the direction of induced current will be such that the magnetic field created by the induced current opposes the changing magnetic field which caused its induction.

The induced emf through

We are given,

The rate of current, di/dt = 0.0260 A/s

The self-induced emf, E = 12.6mV

The magnetic flux, ${\phi }_B$= 0.00285 Wb

The induced EMF in the circuit is given by

$E=L\left|\frac{di}{dt}\right|$

Solve for L we get,

$L=\frac{E}{|di/dt|}$

The self inductance of the solenoid

$L=\frac{N{\phi }_B}{i}$

Through both the equations

$\frac{E}{|di/dt|}=\frac{N{\phi }_B}{i}$

Solving For N

role="math" localid="1664174831050" $$\begin{gathered} N=\frac{E}{N{\phi }_B|di/dt|} \\ =\frac{\left(12.6X{10}^{-3}\right)\left(1.40A\right)}{\left(0.00285WB\right)\left(0.0260A/s\right)} \\ =238turn \\ N=238turn \end{gathered}$$

Therefore, the number of turns that the solenoid possess are 238 turns.