AP Physics B : Using Faraday's Law

Study concepts, example questions & explanations for AP Physics B

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Example Questions

Example Question #1 : Using Faraday's Law

An ideal transformer, under a purely resistive load, consists of an iron core that directs all the magnetic flux from a primary winding through a secondary winding. The primary winding has \displaystyle 1000 closely-packed loops along the coil (such that all see the same magnetic flux). The primary line carries an alternating current at a potential of \displaystyle 10 kV. If the secondary line is to have a potential of \displaystyle 120 V, how many closely-packed loops are required in the secondary winding?

Possible Answers:

\displaystyle 12

\displaystyle 60

\displaystyle 240

\displaystyle 120

\displaystyle 1000

Correct answer:

\displaystyle 12

Explanation:

In basic consideration, the electrical transformer works directly according to Faraday’s law, following the picture given below.

Transformer_figure

If the coils are wound tightly enough that the magnetic flux generated in any of them effectively passes undiminished through all of them, and the iron core directs all the magnetic flux from the primary to secondary coil with insignificant losses, then Faraday’s law points a direct path from \displaystyle V_p to \displaystyle V_s. Faraday’s law relates the time of rate of change of magnetic flux to the voltage (emf) around a closed loop: 

\displaystyle V = \frac{-d \Phi_B}{dt}

If there are \displaystyle N loops stacked together, then the total voltage, \displaystyle V_t, is related to the flux through all loops by:

\displaystyle V_t = -N\frac{d\Phi_B}{dt}

From the figure, it is seen that the voltages in each line are dropped over all loops together. Hence, for the primary winding, by Faraday’s law: 

\displaystyle V_p = -N_p \frac{d \Phi_B}{dt}

For the secondary winding:

\displaystyle V_s = -N_s \frac{d \Phi_B}{dt}

Here, \displaystyle N_p and \displaystyle N_s are the number of coils in the primary and secondary windings. The quantity \displaystyle \frac{d \Phi_B}{dt} is the same in both windings, being transmitted fully (and effectively instantly) by the iron core. This leads to the equation:

\displaystyle \frac{V_p}{N_p} = \frac{V_s}{N_s}

This can be used to solve the problem above:

\displaystyle N_s = \frac{V_s N_p}{V_p} = \frac{(120 V)(1000)}{10,000 V} = 12

 

NOTE:  This only works under alternating current.  Otherwise, there is no change in current to create the changing magnetic flux on which the transformer depends.

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