Question

How many turns of wire must a coil have in order to induce a voltage of 10.5 volts when exposed to a changing magnetic flux with a rate of 0.0075 Wb/s?

Answer 1

Hint: When a coil is exposed to a charging magnetic flux then the imaginary magnetic field lines exert a force on the free electrons and the electrons start to move which results in the form of a flow of current in the coil. The magnitude of the EMF in the coil as defined by Faraday’s second law of electromagnetic induction can be used to solve the above problem.
Faraday’s second law of electromagnetic induction states that the magnitude of emf induced in the coil is equal to the rate of change of flux that linkages with the coil. The flux linkage of the coil is the product of the number of turns in the coil and flux associated with the coil.
Express the formula for the faraday’s second law.
$\therefore E = - N\dfrac{{d\phi }}{{dt}}$ , where $E$ is the induced emf in the coil which is 10.5 volts in the question, $N$ is the number of coils, and $\dfrac{{d\phi }}{{dt}}$ is the rate of change of magnetic flux which is given in the question as 0.0075 Wb/s.
substitute the values.
$\therefore 10.5 = - N \times 0.0075$
$ \Rightarrow - N = \dfrac{{10.5}}{{7.5 \times {{10}^{ - 3}}}}$
$ \Rightarrow N = - 1400$
Here the minus sign is for direction because the formula consists of some vector quantities. The number of turns in a coil cannot be negative, therefore $N = 1400$ . Thus we need 1400 turns in a coil to induce 10.5 voltage exposing to a changing magnetic flux with a rate of 0.0075 Wb/s.

Note: It is clear from the formula that the emf in a coil depends on the number of turns and the rate of changing flux. What is the rate of change of magnetic flux? It is the speed of a magnet bar, say, through the coil. The magnetic bar’s speed is responsible for how much the electrons are moving in a coil. The faster the flux is changing, the more the emf is being induced.